Q59 / 155 · WBJEE 2017 · Mathematics
The common chord of the circles $x^{2}+y^{2}-4 x-4 y=0 \quad$ and $\quad 2 x^{2}+2 y^{2}=32$ subtends at the origin an angle equal to
- A$\frac{\pi}{3}$
- B$\frac{\pi}{4}$
- C$\frac{\pi}{6}$
- Correct answerD$\frac{\pi}{2}$
Solution
Given, equation of circles are $x^{2}+y^{2}-4 x-4 y=0$ and $2 x^{2}+2 y^{2}=32$
or $x^{2}+y^{2}-4 x-4 y=0$
and $x^{2}+y^{2}=16$
$\therefore$ Equation of common chord is
$\left(x^{2}+y^{2}-4 x-4 y\right)-\left(x^{2}+y^{2}-16\right)=0$
$\Rightarrow \quad-4 x-4 y+16=0$
$\Rightarrow \quad x+y=4$
This common chord passes through (2,2)
i.e. centre of first circle. Also, (0,0) is at the circumference of the first circle.
$\therefore$ Common chord will subtent $\frac{\pi}{2}$ angle at (0,0)
Hence, the correct answer is option: 'D'.