Conic-Sections Question 541
Question: The common chord of $ x^{2}+y^{2}-4x-4y=0 $ and $ x^{2}+y^{2}=16 $ subtends at the origin an angle equal to
Options:
A) $ \frac{\pi }{6} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{3} $
D) $ \frac{\pi }{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] The centre of two circles are $ C_1(2,2) $ and $ C_2(0,0) $ . The radii of two circles are $ r_1=2\sqrt{2} $ and $ r_2=4 $ The eq. of the common chord of the circles $ x^{2}+y^{2}-4x-4y=0 $ and $ x^{2}+y^{2}=16 $ is $ x+y=4 $ which meets the circle $ x^{2}+y^{2}=16 $ at points $ A(4,0) $ and $ B(0,4) $ . Obviously $ OA\bot OB $ . Hence, the common chord AB makes a right angle at the centre of the circle $ x^{2}+y^{2}=16. $ Where O is the origin and the centre $ C_2 $ of the second circle.
BETA
