Circles #6 - Questions and Answers

Question 1 To Draw: Reference triangles for given angles on the unit circle. Answer Step 1: If the angle is negative, add 36 0 ∘ 360 ∘ or 2 𝜋 2π radians to convert it to a positive equivalent angle. First circle: Given 𝜃 = − 57 0 ∘ θ=−570 ∘ . 𝜃 = − 57 0 ∘ + 36 0 ∘ = − 21 0 ∘ . θ=−570 ∘ +360 ∘ =−210 ∘ . Since the angle remains negative, add 36 0 ∘ 360 ∘ : 𝜃 = − 21 0 ∘ + 36 0 ∘ = 15 0 ∘ . θ=−210 ∘ +360 ∘ =150 ∘ . Second circle: Given 𝜃 = − 17 𝜋 6   rad θ=− 6 17π ​ rad. Convert to degrees: 𝜃 = − 17 6 ⋅ 18 0 ∘ = − 51 0 ∘ . θ=− 6 17 ​ ⋅180 ∘ =−510 ∘ . Add 36 0 ∘ 360 ∘ : 𝜃 = − 51 0 ∘ + 36 0 ∘ = − 15 0 ∘ . θ=−510 ∘ +360 ∘ =−150 ∘ . Add 36 0 ∘ 360 ∘ again: 𝜃 = − 15 0 ∘ + 36 0 ∘ = 21 0 ∘ . θ=−150 ∘ +360 ∘ =210 ∘ . Step 2: Reference triangles are drawn by plotting the terminal sides for 15 0 ∘ 150 ∘ and 21 0 ∘ 210 ∘ on the unit circle. Label all sides of the triangle, reference angle, and axes. Question 2 Segment 𝐵 𝐶 BC is tangent to circle 𝐴 A at point 𝐵 B. Find the length of segment 𝐵 𝐶 BC. Answer Step 1: Properties of tangents: The tangent at a point of contact is perpendicular to the radius. △ 𝐴 𝐵 𝐶 △ABC is a right triangle. Step 2: Given: 𝐴 𝐵 = 21 AB=21 (radius of the circle). 𝐸 𝐶 = 8 EC=8 and 𝐴 𝐸 = 𝐸 𝐶 = 8 AE=EC=8. Use Pythagorean theorem: 𝐴 𝐶 = 𝐴 𝐸 + 𝐸 𝐶 = 21 + 8 = 29. AC=AE+EC=21+8=29. Apply Pythagorean theorem to △ 𝐴 𝐵 𝐶 △ABC: 𝐴 𝐶 2 = 𝐵 𝐶 2 + 𝐴 𝐵 2 . AC 2 =BC 2 +AB 2 . Substitute values: 2 9 2 = 𝐵 𝐶 2 + 2 1 2 . 29 2 =BC 2 +21 2 . 841 = 𝐵 𝐶 2 + 441. 841=BC 2 +441. 𝐵 𝐶 2 = 841 − 441 = 400. BC 2 =841−441=400. 𝐵 𝐶 = 400 = 20. BC= 400 ​ =20. Final Answer: 𝐵 𝐶 = 20   units BC=20units. Question 3 Find the radius of a circle whose area equals the sum of the areas of two circles with radii 3   in 3in and 4   in 4in. Answer Step 1: Calculate the areas of the two circles. For radius 3 3: 𝐴 1 = 𝜋 𝑟 2 = 𝜋 ( 3 ) 2 = 9 𝜋 . A 1 ​ =πr 2 =π(3) 2 =9π. For radius 4 4: 𝐴 2 = 𝜋 𝑟 2 = 𝜋 ( 4 ) 2 = 16 𝜋 . A 2 ​ =πr 2 =π(4) 2 =16π. Sum of areas: 𝐴 total = 9 𝜋 + 16 𝜋 = 25 𝜋 . A total ​ =9π+16π=25π. Step 2: Let 𝑟 r be the radius of the new circle. Its area is: 𝐴 = 𝜋 𝑟 2 . A=πr 2 . Equating the areas: 𝜋 𝑟 2 = 25 𝜋 . πr 2 =25π. Cancel 𝜋 π: 𝑟 2 = 25. r 2 =25. 𝑟 = 25 = 5. r= 25 ​ =5. Final Answer: 𝑟 = 5   in r=5in. Question 4 To Prove: ∠ 𝐵 𝐴 𝐶 ∠BAC is a right angle ( 9 0 ∘ 90 ∘ ) for two circles tangent externally with a common tangent 𝐵 𝐶 BC. Answer Step 1: Tangent segments drawn from an external point are equal. 𝐷 𝐶 = 𝐴 𝐷 DC=AD, and 𝐴 𝐵 = 𝐵 𝐷 AB=BD. Step 2: Using the property of angles opposite equal sides, we know: ∠ 𝐷 𝐴 𝐵 + ∠ 𝐷 𝐶 𝐴 = 18 0 ∘ ∠DAB+∠DCA=180 ∘ . 2 ∠ 𝐷 𝐴 𝐵 + 2 ∠ 𝐷 𝐴 𝐶 = 18 0 ∘ 2∠DAB+2∠DAC=180 ∘ . Simplifying: ∠ 𝐷 𝐴 𝐵 + ∠ 𝐷 𝐴 𝐶 = 9 0 ∘ . ∠DAB+∠DAC=90 ∘ . Thus, ∠ 𝐵 𝐴 𝐶 = 9 0 ∘ ∠BAC=90 ∘ . Final Answer: ∠ 𝐵 𝐴 𝐶   is a right angle ∠BACis a right angle. Question 5 𝐵 𝐶 BC is a common tangent of circles 𝐴 A and 𝐷 D. Find 𝐸 𝐷 ED and 𝐵 𝐸 BE for given conditions. Answer 1) First Condition: 𝐴 𝐵 = 6   cm , 𝐷 𝐶 = 3   cm , 𝐴 𝐷 = 12   cm AB=6cm,DC=3cm,AD=12cm. Let 𝐸 𝐷 = 𝑥 ED=x. Using similarity of triangles △ 𝐴 𝐸 𝐵 ∼ △ 𝐷 𝐸 𝐶 △AEB∼△DEC: 𝐴 𝐵 𝐷 𝐶 = 𝐴 𝐸 𝐸 𝐷 . DC AB ​ = ED AE ​ . Substitute values: 6 3 = 12 − 𝑥 𝑥 . 3 6 ​ = x 12−x ​ . Simplify: 6 𝑥 = 36 − 3 𝑥 . 6x=36−3x. 9 𝑥 = 36. 9x=36. 𝑥 = 4   cm . x=4cm. 2) Second Condition: 𝐴 𝐵 = 8   cm , 𝐷 𝐶 = 5   cm , 𝐵 𝐶 = 14   cm AB=8cm,DC=5cm,BC=14cm. Let 𝐵 𝐸 = 𝑥 BE=x, and 𝐸 𝐶 = 14 − 𝑥 EC=14−x. Using triangle similarity: 𝐴 𝐵 𝐷 𝐶 = 𝐵 𝐸 𝐸 𝐶 . DC AB ​ = EC BE ​ . Substitute values: 8 5 = 𝑥 14 − 𝑥 . 5 8 ​ = 14−x x ​ . Cross-multiply: 8 ( 14 − 𝑥 ) = 5 𝑥 . 8(14−x)=5x. Simplify: 112 − 8 𝑥 = 5 𝑥 . 112−8x=5x. 112 = 13 𝑥 . 112=13x. 𝑥 = 112 13 . x= 13 112 ​ . Final Answers: 𝐸 𝐷 = 4   cm ED=4cm. 𝐵 𝐸 = 112 13   cm BE= 13 112 ​ cm.