Circle Study Guide

DEFINITION A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point (in the same given plane) remains constant. The fixed point is called the centre of the circle and the constant distance is called the radius of the circle. EQUATION OF A CIRCLE The curve traced by the moving point is called its circumference i.e. the equation of any circle is satisfied by co-ordinates of all points on its circumference. or The equation of the circle means the equation of its circumference. or It is the set of all points lying on the circumference of the circle. Chord and diameter - the line joining any two points on the circumference is called a chord. If any chord passing through its centre is called its diameter. P AB = chord, PQ = diameter A C = centre Standard Equations of the Circle (a) Central Form : If (h, k) is the centre and r is the radius of the circle then its equation is (x–h)2 + (y–k)2 = r2 Q C B r (h, k) Special Cases (i) If centre is origin (0,0) and radius is 'r' then equation of circle is x2 + y2 = r2 and this is called the standard form. (ii) y If radius of circle is zero then equation of circle is (x – h)2 + (y – k)2 = 0. Such circle is called zero circle or point circle. (iii) When circle touches x-axis then equation of the circle is (x–h)2 + (y–k)2 = k2. (iv) (v) C k 0 Touching x-axis y x C (h,k) h When circle touches y-axis then equation of circle is (x–h)2 + (y–k)2 = h2 . When circle touches both the axes (x-axis and y-axis) then equation of circle (x–h)2 + (y–h)2 = h2. (h,k) 0 Touching y-axis x y C (h,h) h h 0 Touching x-axis and y-axis x CIRCLE (vi) When circle passes through the origin and centre of the circle is (h,k) then radius h 2  k 2  r and intercept cut on x-axis OP =2h, y Q (0,2k) and intercept cut on y-axis is OQ = 2k and equation of circle is (x–h)2 + (y–k)2 = h2 + k2 or x2 + y2 – 2hx – 2ky = 0 (b) C (h,k) k O P (2h,0) x General Equation of Circle x2 + y2 + 2gx + 2fy + c = 0. where g,f,c are constants and centre is (–g,–f)  coefficient of x coefficient of y  , i.e.     2 2 2 2 and radius r  g  f  c (i) If (g2 + f2 – c) > 0, then r is real and positive and the circle is a real circle. (ii) If (g2 + f2 – c) = 0, then radius r = 0 and circle is a point circle. (iii) If (g2 + f2 –c)<0, then r is imaginary then circle is also an imaginary circle with real centre. (iv) x2 + y2 + 2gx + 2fy + c = 0, has three constants and to get the equation of the circle at least three conditions should be known (v) (c)  A unique circle passes through three non collinear points. The general second degree in x and y, ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 represents a circle if : • coefficient of x2 = coefficient of y2 or a = b  0 • coefficient of xy = 0 or • (g2 + f2 – c)  0 (for a real circle) h=0 Equation of Circle in Diameter Form : If A(x1,y1) and B(x2,y2) are the end points of the diameter of the circle and P(x,y) is the point other then A and B on the circle then from geometry we know that APB = 90°.  (Slope of PA) × (Slope of PB) = –1   y  y1  y  y2    x  x1  x  x2  (x–x1) (x–x2)+(y–y1)(y–y2) = 0 P(x,y) (x1 ,y1) A C B(x2,y2)  =–1  This will be the circle of least radius passing through (x1, y1) and (x2, y2). (d) Equation of Circle in Parametric Forms : (i) The parametric equation of the circle x2+y2 = r2 are x = r cos, y = r sin ;   [0, 2) and (r cos , r sin ) are (ii) called the parametric co-ordinates. The parametric equation of the circle (x – h)2 + (y – k)2 = r2 is x = h + rcos, y = k + rsin where  is parameter. (iii) The parametric equation of the circle x2 + y2 + 2gx + 2fy + c = 0 are x = – g + y = –f + g 2 + f 2 – c sin  where  is parameter.. g 2 + f 2 – c cos, ETOOS KEY POINTS Equation of a straight line joining two point  on the circle x2 + y2 = a2 is x cos       + y sin = a cos . 2 2 2 Ex. Find the equation of the circle whose centre is (1, –2) and radius is 4. Sol. The equation of the circle is (x – 1)2 + (y – (–2))2 = 42  (x – 1)2 + (y + 2)2 = 16  x2 + y2 – 2x + 4y – 11 = 0 Ex. Find the equation of the circle, the coordinates of the end points of whose diameter are (–1, 2) and (4, –3) Sol. We know that the equation of the circle described on the line segment joining (x1, y1) and (x2, y2) as a diameter is (x – x1) (x – x2) + (y – y1) (y – y2) = 0. Here, x1 = –1, x2 = 4, y1 = 2 and y2 = –3. So, the equation of the required circle is (x + 1) (x – 4) + (y – 2) (y + 3) = 0  x2 + y2 – 3x + y – 10 = 0. Ex. If y = 2x + m is a diameter to the circle x2 + y2 + 3x + 4y – 1 = 0, then find m Sol. Centre of circle = (–3/2 , –2). This lies on diameter y = 2x + m  – 2 = (–3/2) × 2 + m  m=1 Ex. A circle has radius equal to 3 units and its centre lies on the line y = x – 1. Find the equation of the circle if it passes through (7, 3). Sol. Let the centre of the circle be (). It lies on the line y = x – 1   – 1. Hence the centre is ( –1).  The equation of the circle is (x – )2 + (y –  + 1)2 = 9 It passes through (7, 3)  (7 – )2 + (4 – )2 = 9  22 – 22 + 56 = 0  2 – 11 + 28 = 0  ( – 4)( – 7) = 0   = 4, 7 Hence the required equations are x2 + y2 – 8x – 6y + 16 = 0 and x2 + y2 – 14x – 12y + 76 = 0. Ex. Find the centre & radius of the circle whose equation is x2 + y2 – 4x + 6y + 12 = 0 Sol. Comparing it with the general equation x2 + y2 + 2gx + 2fy + c = 0, we have 2g = – 4  g = –2 2f = 6  f=3 &  c = 12 centre is (–g, –f) i.e. (2, –3) and radius = g2  f 2  c = ( 2) 2  (3) 2  12 = 1 CIRCLE INTERCEPTS CUT BY THE CIRCLE ON AXES The intercepts cut by the circle x2 + y2 + 2gx + 2fy + c =0 on : (i) (ii) x-axis = 2 g 2 – c y-axis = 2 f 2 – c (i) If the circle cuts the x-axis at two distinct point, then g2 – c > 0 (ii) If the cirlce cuts the y-axis at two distinct point, then f2 – c > 0 (iii) If circle touches x-axis then g2 = c. O (iv) If circle touches y-axis then f2 = c. a P (v) Circle lies completely above or below the x-axis then g2 < c. B A C (vi) Circle lies completely to the right or left to the y-axis, then f2 < c. (vii) Intercept cut by a line on the circle x2 + y2 + 2gx + 2fy+c=0 or length of chord of the circle  2 a 2  P 2 where a is the radius and P is the length of perpendicular from the centre to the chord. Ex. Find the equation to the circle touching the negative y-axis at a distance 3 from the origin and intercepting a length 8 on the x-axis. Sol. Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0. Since it touches y-axis at (0, –3) and (0, – 3) lies on the circle.  ...... (i) c = f2 ...... (ii) 9 – 6f + c = 0 From (i) and (ii), we get 9 – 6f + f = 0 2  (f – 3)2 = 0  f = 3. Putting f = 3 in (i) we obtain c = 9. It is given that the circle x2 + y2 + 2gx + 2fy + c = 0 intercepts length 8 on x-axis  2 g2  c = 8  2 g2  9 = 8  g2 – 9 = 16  g=±5 Hence, the required circle is x2 + y2 ± 10x + 6y + 9 = 0. BASIC THEOREMS & RESULTS OF CIRCLES (a) (c) Concentric circles : Circles having same centre. (b) Congruent circles : Iff their radii are equal. Congruent arcs : Iff they have same degree measure at the centre. Theorem 1 (i) If two arcs of a circle (or of congruent circles) are congruent, the corresponding chords are equal. Converse : If two chords of a circle are equal then their corresponding arcs are congruent. (ii) Theorem 2 (i) Equal chords of a circle (or of congruent circles) subtend equal angles at the centre. Converse : If the angle subtended by two chords of a circle (or of congruent circles) at the centre are equal, the chords are equal. (ii) The perpendicular from the centre of a circle to a chord bisects the chord. Converse : The line joining the mid point of a chord to the centre of a circle is perpendicular to the chord. Perpendicular bisectors of two chords of a circle intersect at its centre. (i) (ii) There is one and only one circle passing through three non collinear points. If two circles intersects in two points, then the line joining the centres is perpendicular bisector of common chords. Theorem 3 Theorem 4 (i) (ii) (iii) Theorem 5 (i) (ii) (iii) Equal chords of a circle (or of congruent circles) are equidistant from the centre. Converse : Chords of a circle (or of congruent circles) which are equidistant from the centre are equal in length. If two equal chords are drawn from a point on the circle, then the centre of circle will lie on angle bisector of these two chords. Of any two chords of a circle larger will be near to centre. The degree measure of an arc or angle subtended by an arc at the centre is double the angle subtended by it at any point of alternate segment. Angle in the same segment of a circle are equal. The angle in a semi circle is right angle. Converse : The arc of a circle subtending a right angle in alternate segment is semi circle. Theorem 6 Any angle subtended by a minor arc in the alternate segment is acute and any angle subtended by a major arc in the alternate segment is obtuse. Theorem 7 If a line segment joining two points subtends equal angles at two other points lying on the same side of the line segment, the four points are concyclic, i.e. lie on the same circle. (d) Cyclic Quadrilaterals A quadrilateral is called a cyclic quadrilateral if its all vertices lie on a circle. Theorem 1 A D B C The sum of either pair of opposite angles of a cyclic quadrilateral is 180° OR The opposite angles of a cyclic quadrilateral are supplementary. Converse : If the sum of any pair of opposite angle of a quadrilateral is 180°, then the quadrilateral is cyclic. Theorem 2 If a side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle. Theorem 3 The internal angle bisectors of a cyclic quadrilateral form a quadrilateral which is also cyclic. C D Q A P R S B CIRCLE Theorem 4 If two sides of a cyclic quadrilateral are parallel then the remaining two sides are equal and the diagonals are also equal. OR A cyclic trapezium is isosceles and its diagonals are equal. Converse : If two non-parallel sides of a trapezium are equal, then it is cyclic. OR An isosceles trapezium is always cyclic. Theorem 5 When the opposite sides of cyclic quadrilateral (provided that they are not parallel) are produced, then the exterior angle bisectors intersect at right angle. POSITION OF A POINT WITH RESPECT TO A CIRCLE The point (x1, y1) is inside, on or outside the circle S  x2 + y2 + 2gx + 2fy + c = 0. according as S1  x1² + y1² + 2gx1 + 2fy1 + c <, = or > 0. The greatest & the least distance of a point A from a circle with centre C & radius r is AC + r & |AC – r| respectively. Ex. Discuss the position of the points (1, 2) and (6, 0) with respect to the circle x2 + y2 – 4x + 2y – 11 = 0 Sol. We have x2 + y2 – 4x + 2y – 11 = 0 or S = 0, where S = x2 + y2 – 4x + 2y – 11. For the point (1, 2), we have S1 = 12 + 22 – 4 × 1 +2 × 2 – 11 < 0 For the point (6, 0), we have S2 = 62 + 02 – 4 × 6 +2 × 0 – 11 > 0 Hence, the point (1, 2) lies inside the circle and the point (6, 0) lies outside the circle. LINE AND A CIRCLE Let L = 0 be a line & S = 0 be a circle. If r is the radius of the circle & p is the length of the perpendicular from the centre on the line, then: (i) p>r  the line does not meet the circle i. e. passes out side the circle. (ii) p=r  the line touches the circle. (It is tangent to the circle) (iii) p a (1 + m )  the line does not meet the circle i. e. passes out side the circle. 2 2 2 2 2 2 C P r nt ca Se t gen Tan These conditions can also be obtained by solving y = mx + c with x2 + y2 = a2 and making the discriminant of the quadratic greater than zero for secant, equal to zero for tangent and less the zero for the last case. Ex. For what value of c will the line y = 2x + c be a tangent to the circle x2 + y2 = 5 ? Sol. We have y = 2x + c or 2x – y + c = 0 ......(i) and x 2 + y2 = 5 ........(ii) If the line (i) touches the circle (ii), then length of the  from the centre (0, 0) = radius of circle (ii)   20  0  c 2 2  (1) c =± 5 2 5 = 5  c = 5  c=±5 5 Hence, the line (i) touches the circle (ii) for c = ± 5 TANGENT LINE OF CIRCLE When a straight line meet a circle on two coincident points then it is called the tangent of the circle. (a) Condition of Tangency : (P>r) The line L = 0 touches the circle S = 0 if P the length of the (P=r) (P 0 Hence point P lies outside the circle For point P (1, 2), T = x . 1 + y . 2 – (x + 1) + 2(y + 2) + 7 i.e. T = 4y + 10 ...... (i) Now equation of the chord of contact of point P(1, 2) w.r.t. circle (i) will be 4y + 10 = 0 or 2y + 5 = 0 Ex. The chord of contact of tangents drawn from a point on the circle x2 + y2 = a2 to the circle x2 + y2 = b2 touches the circle x2 + y2 = c2. Show that a, b, c are in GP. Sol. Let P(acos, asin) be a point on the circle x2 + y2 = a2. Then equation of chord of contact of tangents drawn from P(acos, asin) to the circle x2 + y2 = b2 is axcos + aysin = b2 2 2 This touches the circle x + y = c   or 2 | 0  0  b2 | (a 2 cos 2   a 2 sin 2 ) b2 = ac  ..... (i) ..... (ii) Length of perpendicular from (0, 0) to (i) = radius of (ii) c a, b, c are in GP. T P R x2 + y2= c2 2 2 x2 + y = b x2+ y2= a2 EQUATION OF THE CHORD WITH A GIVEN MIDDLE POINT (T = S1) The equation of the chord of the circle S  x2 + y2 + 2gx + 2fy + c = 0 in terms of its mid point M (x1 , y1) is x1  g (x  x1). This on simplification can be put in the form y1  f xx1 + yy1 + g (x + x1) + f (y + y1) + c = x12 + y12 + 2gx1 + 2fy1 + c which is designated by T = S1. y  y1 =  C • A• • y 1) , M(x 1 •B (i) The shortest chord of a circle passing through a point ‘M’ inside the circle , is one chord whose middle point is M. (ii) The chord passing through a point ' M ' inside the circle and which is at a maximum distance from the centre is a chord with middle point M. Ex. Find the equation of the chord of the circle x2 + y2 + 6x + 8y – 11 = 0, whose middle point is (1, –1) Sol. Equation of given circle is S  x2 + y2 + 6x + 8y – 11 = 0 Let L  (1, –1) For point L(1, –1), S1 = 12 + (–1)2 + 6.1 + 8(–1) – 11 = –11 and T  x.1 + y (–1) + 3(x + 1) + 4(y – 1) – 11 i.e. T  4x + 3y – 12 Now equation of the chord of circle (i) whose middle point is L(1, –1) is T = S1 or 4x + 3y – 12 = –11 or 4x + 3y – 1 = 0 Second Method : Let C be the centre of the given circle, then C  (–3, –4). L  (1, –1) slope of CL =  Equation of chord of circle whose middle point is L, is y + 1 = – ( chord is perpendicular to CL) or 4x + 3y – 1 = 0 4 (x – 1) 3 4  1 3 = 3  1 4 Ex. Find the locus of middle points of chords of the circle x2 + y2 = a2, which subtend right angle at the point (c, 0). Sol. Let N(h, k) be the middle point of any chord AB, y A which subtend a right angle at P(c, 0). Since APB = 90°  NA = NB = NP x' (h,k) N P(c,0) (since distance of the vertices from middle point of B the hypotenuse are equal) or (NA)2 = (NB)2 = (h – c)2 + (k – 0)2 ..... (i) O y' But also BNO = 90°  (OB)2 = (ON)2 + (NB)2  –(NB)2 = (ON)2 – (OB)2 or 2(h2 + k2) – 2ch + c2 – a2 = 0  Locus of N(h, k) is 2(x2 + y2) – 2cx + c2 – a2 = 0  –[(h – c)2 + (k – 0)2] = (h2 + k2) – a2 x CIRCLE EQUATION OF THE CHORD JOINING TWO POINTS OF CIRCLE The equation of chord PQ to the circle x2 + y2 = a 2 joining two points P() and Q() on it is given by the equation of a straight line joining two point  &  on the circle x2 + y2 = a 2 is x cos      + y sin = a cos . 2 2 2 DIRECT AND TRANSVERSE COMMON TANGENTS Let two circles having centre C1 and C2 and radii, r1 and r2 and C1C2 is the distance between their centres then : (a) Both Circles will Touch (i) (ii) (b) Externally : if C1C2 = r1 + r2 i.e. the distance between their centres is equal to sum of their radii and point P & T divides C1C2 in the ratio r1 : r2 (internally & externally respectively). In this case there are three common tangents. C1 P T C2 Internally : if C1C2 = |r1–r2| i.e. the distance between their centres is equal to difference between their radii and point P divides C1C2 in the ratio r1 : r2 externally and in this case there will be only one common tangent. C1 C2 P The Circles will Intersect when |r1 – r2| < C1C2 < r1 + r2 in this case there are C1 two common tangents. (c) C2 The Circles will not Intersect (i) One circle will lie inside the other circle if C1C2 < | r1–r2| In this case there will be no common tangent. (ii) When circle are apart from each other then C1C2> r1 + r2 and in this case there will be four common tangents. Lines PQ and RS are called transverse or indirect or internal common tangents and these lines meet line C1C2 on T1and T1 divides the line C1C2 in the ratio r1 : r2 internally and lines AB & CD are called direct or external common tangents. These lines meet C1C2 produced on T2. Thus T2 divides C1C2 externally in the ratio r1 : r2. R A Q B C1 C P T1 C2 D S (i) The direct common tangents meet at a point which divides the line joining centre of circles externally in the ratio of their radii. Transverse common tangents meet at a point which divides the line joining centre of circles internally in the ratio of their radii. (ii) Length of an external (or direct) common tangent & internal (or transverse) common tangent to the two circles are given by: Lext = d 2  (r1  r2 )2 & Lint = d 2  (r1  r2 )2 , where d = distance between the centres of the two circles and r 1, r 2 are the radii of the two circles. Note that length of internal common tangent is always less than the length of the external or direct common tangent. T2 Ex. Examine if the two circles x2 + y2 – 2x – 4y = 0 and x2 + y2 – 8y – 4 = 0 touch each other externally or internally. Sol. Given circles are x2 + y2 – 2x – 4y = 0 ...... (i) and x + y – 8y – 4 = 0 ...... (ii) 2 2 Let A and B be the centres and r1 and r2 the radii of circles (i) and (ii) respectively, then A  (1, 2), B  (0, 4), r1 = 5, r2 = 25 Now AB = (1  0) 2  (2  4) 2 = 5 and r 1 + r2 = 3 5 , |r 1 – r2| = 5 Thus AB = |r1 – r2|, hence the two circles touch each other internally. THE ANGLE OF INTERSECTION OF TWO CIRCLES Definition : The angle between the tangents of two circles at the point of intersection of the two circles is called angle of intersection of two circles. If two circles are S1  x2 + y2 + 2g1x + 2f1y + c1 = 0 S2  x2 + y2 + 2g2x + 2f2y + c2 = 0 and  is the acute angle between them then cos   2 g1 g 2  2 f1 f 2  c1  c2 2 g12  f12  c1 or g 22  f 22  c2 P  r 2  r22  d 2  cos    1  2r1r2   r1 C1 r2 d C2 Here r1 and r2 are the radii of the circles and d is the distance between their centres. If the angle of intersection of the two circles is a right angle then such circles are called "Orthogonal circles" and conditions for the circles to be orthogonal is 2g1g2 + 2f1f2 = c1+ c2 (a) The centre of a variable circle orthogonal to two fixed circles lies on the radical axis of two circles. (b) If two circles are orthogonal, then the polar of a point 'P' on first circle w.r.t. the second circle passes through the point Q which is the other end of the diameter through P. Hence locus of a point which moves such that its polars w.r.t. the circles S1 = 0 , S2 = 0 & S3 = 0 are concurrent in a circle which is orthogonal to all the three circles. (c) The centre of a circle which is orthogonal to three given circles is the radical centre provided the radical centre lies outside all the three circles. Ex. Sol. Obtain the equation of the circle orthogonal to both the circles x 2 + y 2 + 3x – 5y+ 6 = 0 and 4x 2 + 4y 2 – 28x +29 = 0 and whose centre lies on the line 3x + 4y + 1 = 0. Given circles are x2 + y2 + 3x – 5y + 6 = 0 ...... (i) and 4x2 + 4y2 – 28x + 29 = 0 29 or x2 + y2 – 7x + = 0. ...... (ii) 4 2 2 Let the required circle be x + y + 2gx + 2fy + c = 0 ...... (iii) Since circle (iii) cuts circles (i) and (ii) orthogonally  3  5 2g   + 2f    = c + 6 2    2 or 3g – 5f = c + 6 ...... (iv) and  7 29 2g    + 2f.0 = c + 4  2 or – 7g = c + 29 4 ...... (v) CIRCLE (iv) & (v), we get 10g – 5f = – 5 4 or 40g – 20f = – 5. Given line is 3x + 4y = – 1 Since centre (– g, – f) of circle (iii) lies on line (vii),  – 3g – 4f = – 1 1 Solving (vi) & (viii), we get g = 0, f = 4   ...... (vi) ...... (vii) ...... (viii) 29 4 from (iii), required circle is from (5), c = – x 2 + y2 + 1 29 y– =0 2 4 or 4(x2 + y2) + 2y – 29 = 0 POLE AND POLAR Let any straight line through the given point A(x1,y1) intersect the given circle S =0 in two points P and Q and if the tangent of the circle at P and Q meet at the point R then locus of point R is called polar of the point A and point A is called the pole, with respect to the given circle. (a) R (h,k) A (x 1,y 1) The Equation of the Polar of Point (x1,y1) w.r.t. Circle x2 + y2 = a2 (T = 0). S (h,k) Let PQR is a chord which passes through the point P(x1,y1) which intersects the circle at points Q and R and the tangents are drawn at points Q and R meet at point S(h,k) then equation of QR the chord of contact is x1h + y1k= a 2 Q P (x1,y1) P Q  locus of point S(h,k) is xx1 + yy1 = a2 which is the equation of the polar. (i) The equation of the polar is the T=0, so the polar of point (x1,y1) w.r.t circle x2 + y2 + 2gx + 2fy + c = 0 is xx1+ yy1+ g(x + x1) + f(y + y1)+c = 0 (ii) If point is outside the circle then equation of polar and chord of contact is same. So the chord of contact is polar. (iii) If point is inside the circle then chord of contact does not exist but polar exists. (iv) If point lies on the circle then polar , chord of contact and tangent on that point are same. (v) If the polar of P w.r.t. a circle passes through the point Q, then the polar of point Q will pass through P and hence P Q are conjugate points of each other w.r.t. the given circle. (vi) If pole of a line w.r.t. a circle lies on second line. Then pole of second line lies on first line and hence both lines are conjugate lines of each other w.r.t. the given circle. (vii) If O be the centre of a circle and P be any point, then OP is perpendicular to the polar of P. (viii) If O be the centre of a circle and P any point, then if OP (produce, if necessary) meet the polar of P in Q, then OP. OQ = (radius)2 R (b) Pole of a Given Line with Respect to a Circle To find the pole of a line we assume the coordinates of the pole then from these coordinates we find the polar. This polar and given line represent the same line. Then by comparing the coefficients of similar terms we can get the coordinates of the pole. The pole of x + my + n = 0  l a 2 ma 2  , w.r.t. circle x2 + y2 = a2 will be   n   n Ex. Find the equation of the polar of the point (2, –1) with respect to the circle x2 + y2 – 3x + 4y – 8 = 0 Sol. Given circle is x2 + y2 – 3x + 4y – 8 = 0 ...... (i) Given point is (2, –1) let P = (2, –1). Now equation of the polar of point P with respect to circle (i)  x2  y 1  x.2 + y(–1) – 3   +4   –8=0  2   2  or 4x – 2y – 3x – 6 + 4y – 4 – 16 = 0 or x + 2y – 26 = 0 RADICAL AXIS OF THE TWO CIRCLES (S1 – S2= 0) P(h,k) Definition The locus of a point, which moves in such a way that the length of tangents drawn from it to the circles are equal and is called the radical axis. If two circles are S1  x2 + y2 + 2g1x + 2f1 y + c1 =0 S2  x2 + y2 + 2g2x + 2f2y + c2 = 0 A B Radical axis Let P(h,k) is a point and PA,PB are length of two tangents on the circles from point P, Then from definition - h 2  k 2  2 g1h  2 f1k  c1  h 2  k 2  2 g 2 h  2 f 2 k  c2 or 2(g1–g2) h + 2(f1–f2) k + c1 – c2 = 0 locus of (h,k) 2x(g1–g2) + 2y(f1–f2)k + c1 – c2 = 0 S1– S2= 0 which is the equation of radical axis.  (i) To get the equation of the radical axis first of all make the coefficient of x2 and y2 =1 (ii) If circles touch each other then radical axis is the common tangent to both the circles. (iii) When the two circles intersect on real points then common chord is the radical axis of the two circles. (iv) The radical axis of the two circles is perpendicular to the line joining the centre of two circles but not always pass through mid point of it. (v) Radical axis (if exist) bisects common tangent to two circles. (vi) The radical axes of three circles (taking two at a time) meet at a point. (vii) If circles are concentric then the radical axis does not always exist but if circles are not concentric then radical axis always exists. (viii) If two circles are orthogonal to the third circle then radical axis of both circle passes through the centre of the third circle. (ix) A system of circle, every pair of which have the same radical axis, is called a coaxial system of circles. CIRCLE RADICAL CENTRE The radical centre of three circles is the point from which length of tangents on three circles are equal i.e. the point of intersection of radical axis of the circles is the radical centre of the circles. To get the radical axis of three circles S1 =0, S2=0, S3=0 we have to solve any two S1–S2=0, S2–S3=0, S3–S1=0 I T1 = C1 = C2 = T2 III C3 T3 II (i) The circle with centre as radical centre and radius equal to the length of tangent from radical centre to any of the circle, will cut the three circles orthogonally. (ii) If three circles are drawn on three sides of a triangle taking them as diameter then its orthocenter will be its radical centre. (iii) Locus of the centre of a variable circle orthogonal to two fixed circles is the radical axis between the two fixed circles. (iv) If two circles are orthogonal, then the polar of a point 'P' on first circle w.r.t. the second circle passes through the point Q which is the other end of the diameter through P. Hence locus of a point which moves such that its polars w.r.t. the circles S1 = 0 , S2 = 0 & S3 = 0 are concurrent is a circle which is orthogonal to all the three circles. Ex. Find the co-ordinates of the point from which the lengths of the tangents to the following three circles be equal. 3x2 + 3y2 + 4x – 6y – 1 = 0 2x2 + 2y2 – 3x – 2y – 4 = 0 2x2 + 2y2 – x + y – 1 = 0 Sol. Here we have to find the radical centre of the three circles. First reduce them to standard form in which coefficients of x2 and y2 be each unity. Subtracting in pairs the three radical axes are 17 5 x–y+ =0; 6 3 – –x– 3 3 y– =0 2 2 11 5 1 x + y – = 0. 6 2 6  16 31  solving any two, we get the point     which satisfies the third also. This point is called the radical  21 63  centre and by definition the length of the tangents from it to the three circles are equal. Ex. A and B are two fixed points and P moves such that PA = nPB where n  1. Show that locus of P is a circle and for different values of n all the circles have a common radical axis. Sol. Let A  (a, 0), B  (–a, 0) and P(h, k) So PA = nPB  (h – a)2 + k2 = n2[(h + a)2 + k2]  (1 – n2)h2 + (1 – n2)k2 – 2ah(1 + n2) + (1 – n2)a2 = 0  1  n2   a2  0 h2 + k2 – 2ah  2   1n Hence locus of P is   1  n2  2 x2 + y2 – 2ax    a  0 , which is a circle of different values of n. 1  n2  Let n1 and n2 are two different values of n so their radical axis is x = 0 i.e. y-axis. Hence for different values of n the circles have a common radical axis. FAMILY OF CIRCLES (a) (b) (c) The equation of the family of circles passing through the points of intersection of two circles S1 = 0 & S2 = 0 is : S1 + K S2 = 0 (K  –1). The equation of the family of circles passing through the point of intersection of a circle S = 0 & a line L = 0 is given by S + KL = 0. S1 S2 S The equation of a family of circles passing through two given points (x1 , y1) & (x2 , y2) can be written in the form : x (x  x1) (x  x2) + (y  y1) (y  y2) + K x1 x2 (d) L y y1 1 1 = 0 where K is a parameter.. 1 y2 C A (x1 ,y1 ) The equation of a family of circles touching a fixed line y  y1 = m (x  x1) B (x2,y2) (x1 ,y1 ) at the fixed point (x1 , y1) is (x  x1) + (y  y1) + K [y  y1  m (x  x1)] = 0, 2 2 where K is a parameter. (e) Family of circles circumscribing a triangle whose sides are given by 1 L1 = 0 ; L2 = 0 & L3 = 0 is given by ; L1L2 +  L2L3 +  L3L1 = 0 provided coefficient of xy = 0 & coefficient of x2 = coefficient of y2. (f) 3 Equation of circle circumscribing a quadrilateral whose sides in 2 order are represented by the lines L1 = 0, L2 = 0, L3 = 0 & L4 = 0 is 1 L1L3 + L2L4 = 0 provided coefficient of x2 = coefficient of y2 and coefficient of xy = 0. Ex. Find the equations of the circles passing through the points of intersection of the circles x2 + y2 –2x – 4y – 4 = 0 and x2 + y2 – 10x – 12y + 40 = 0 and whose radius is 4. Sol. Any circle through the intersection of given circles is S1 + S2 = 0 or (x2 + y2 – 2x – 4y – 4) + (x2 + y2 – 10x – 12y + 40 ) = 0 or (x2 + y2) – 2 (1  5) (2  6) 40  4 x–2 y+ =0 1  1  1  2 ...........(i) 3 4 CIRCLE r=  g 2  f 2  c = 4, given 16 = (1  5 ) 2 (2  6 ) 2 40  4 + – (1  ) 2 (1   ) 2 1  16(1 + 2 + 2) = 1 + 10 + 252 + 4 + 24 + 362 – 402 – 40 + 4 + 4 or 16 + 32 + 162 = 212 – 2 + 9 or 52 – 34 – 7 = 0  ( – 7) (5 + 1) = 0   = 7, – 1/5 Putting the values of  in (i) the required circles are 2x2 + 2y2 – 18x – 22y + 69 = 0 and x2 + y2 – 2y – 15 = 0 Ex. Find the equation of circle circumscribing the triangle whose sides are 3x – y – 9 = 0, 5x – 3y – 23 = 0 & x + y – 3 = 0. Sol. A L1: 3x – y – 9 = 0 B L2: 5x – 3y –23 = 0 L3: x + y – 3 = 0 C L1L2 + L2L3 + µL1L3 = 0 (3x – y – 9) (5x – 3y – 23) + (5x – 3y – 23) (x + y – 3) + µ (3x – y – 9) (x + y – 3) = 0 (15x2 + 3y2 – 14xy – 114x + 50y + 207) + (5x2 – 3y2 + 2xy – 38x – 14y + 69) + µ (3x2 – y2 + 2xy – 18x – 6y + 27) = 0 (5 + 3µ + 15)x2 + (3 – 3 – µ)y2 + xy (2 + 2µ – 14) – x (114 + 38 + 18µ) + y(50 – 14 – 6µ) + (207 + 69 + 27µ) = 0 ...... (i) coefficient of x2 = coefficient of y2  5 + 3µ + 15 = 3 – 3 – µ 2 + µ + 3 = 0 ...... (ii) coefficient of xy = 0  + µ – 7 = 0 Solving (ii) and (iii), we have  = – 10, µ = 17 Putting these values of  & µ in equation (i), we get 2x2 + 2y2 – 5x + 11y – 3 = 0 ...... (iii) 1. Definition A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point (in the same given plane) remains constant. The fixed point is called the centre of the circle and the constant distance is called the radius of the circle. 2. Standard Equations of the Circle (a) Central Form If (h, k) is the centre and r is the radius of the circle then its equation is (x–h)2 + (y–k)2 = r2 (b) General Equation of Circle x2 + y2 + 2gx + 2fy + c = 0. where g,f,c are constants and centre is (–g,–f)  coefficient of x coefficient of y  , i.e.     2 2 2 2 and radius r  g  f  c Note The general quadratic equation in x and y, ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 represents a circle if : (c) (i) coefficient of x2 = coefficient of y2 or a = b  0. (ii) coefficient of xy = 0 or h = 0 (iii) (g2 + f 2 – c)  0 (for a real circle) Intercepts Cut by the Circle on Axes The intercepts cut by the circle x2 + y2 + 2gx + 2fy + c =0 on : (i) Note x-axis = 2 g 2 – c (ii) y-axis = 2 f2 – c Intercept cut by a line on the circle x2 + y2 + 2gx + 2fy+c=0 or length of chord of the circle  2 a 2  P 2 where a is the radius and P is the length of perpendicular from the centre to the chord. (d) Equation of Circle in Diameter Form If A(x1,y1) and B(x2,y2) are the end points of the diameter of the circle then the equation of circle is given by  (e) P(x,y) (x1,y1) A (x – x1) (x – x2) + (y – y1)(y – y2) = 0 C B(x2,y2) Equation of Circle in Parametric Forms (i) (ii) The parametric equation of the circle x2+y2 = r2 are x = r cos, y = r sin ;   [0, 2) and (r cos , r sin ) are called the parametric co-ordinates. The parametric equation of the circle (x – h)2 + (y – k)2 = r2 is x = h + r cos, y = k + r sin  where  is parameter. (iii) The parametric equation of the circle x2 + y2 + 2gx + 2fy + c = 0 are x = – g + 2 2 y = –f + g + f – c sin  where  is parameter.. Note Equation of a straight line joining two point  on the circle x2 + y2 = a2 is x cos       + y sin = a cos . 2 2 2 g 2 + f 2 – c cos, CIRCLE 3. Position of a Point W.R.T Circle (a) Let the circle is x2 + y2 + 2gx + 2fy + c = 0 and the point is (x1,y1) then Point (x 1,y 1) lies out side the circle or on the circle or inside the circle according as C A (b) 4. (x1,y1) Q  x12 + y12 + 2gx1 +2fy1 + c >, =, < 0 or S1 >, =, < 0 The greatest & the least distance of a point A from a circle with centre C & radius r is AC + r & |AC – r| respectively. Tangent Line of Circle When a straight line meet a circle on two coincident points then it is called the tangent of the circle. (a) Condition of Tangency (P>r) (P=r) (Pr1+r2 and in this case there will be four common tangents. Lines PQ and RS are called transverse or indirect or internal common tangents and these lines meet line C1C2 on T1and T1 divides the line C1C2 in the ratio r1 : r2 internally and lines AB & CD are called direct or external common tangents. These lines meet C1C2 produced on T2. Thus T2 divides C1C2 externally in the ratio r1 : r2. Note Length of direct common tangent =  r2 )2 Length of transverse common tangent = (C 1 C 2 )2  (r1  r2 )2 12. The Angle of Intersection of Two Circles Definition : The angle between the tangents of two circles at the point of intersection of the two circles is called angle of intersection of two circles. If two circles are S1  x2 + y2 + 2g1x + 2f1y + c1 = 0 S2  x2 + y2 + 2g2x + 2f2y + c2 = 0 and  is the acute angle between them CIRCLE then cos   2 g1 g 2  2 f1 f2  c1  c 2 2 g12  f12  c1 g22  f22  c 2 or  r 2  r22  d 2  cos    1  2r1 r2   P r1 C1 r2 d C2 Here r1 and r2 are the radii of the circles and d is the distance between their centres. If the angle of intersection of the two circles is a right angle then such circles are called "Orthogonal circles" and conditions for the circles to be orthogonal is -2g1g2 + 2f1f2 = c1+ c2 13. Radical axis of the Two Circles (S1 – S2= 0) Definition : The locus of a point, which moves in such a way that the length of tangents drawn from it to the circles are equal and is called the radical axis. If two circles are - P(h,k) A B Radical axis S1  x2 + y2 + 2g1x + 2f1 y + c1 =0 S2  x2 + y2 + 2g2x + 2f2y + c2 = 0 Let P(h,k) is a point and PA,PB are length of two tangents on the circles from point P, Then from definition - h 2  k 2  2 g1 h  2 f1 k  c1  h 2  k 2  2 g2 h  2 f2 k  c 2 or 2(g1–g2) h + 2(f1–f2) k + c1 – c2 = 0  locus of (h,k) 2x(g1–g2) + 2y(f1–f2)k + c1 – c2 = 0 S1– S2= 0 Note (i) If two circles touches each other then common tangent is radical axis. S1 – S2 = 0 S1 – S2 = 0 (ii) If two circle cuts each other then common chord is radical axis. (iii) If two circles cuts third circle orthogonally then radical axis of first two is locus of centre of third circle. S1 – S2 = 0