question_answer

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

A) \[\frac{\pi }{6}\]

B) \[\frac{\pi }{4}\]

C) \[\frac{\pi }{3}\]

D) \[\frac{\pi }{2}\]

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.