Question:
The common chord of the circles $x^2 + y^2 - 4x - 4y = 0$ and $2x^2 + 2y^2 = 32$ subtends at the origin an angle equal to
The common chord of the circles $x^2 + y^2 - 4x - 4y = 0$ and $2x^2 + 2y^2 = 32$ subtends at the origin an angle equal to
Updated On: June 02, 2025
- $\frac{\pi}{3}$
- $\frac{\pi}{4}$
- $\frac{\pi}{6}$
- $\frac{\pi}{2}$
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The Correct Option is D
Solution and Explanation
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 $
$\Rightarrow -4 x-4 y+16=0$
$\Rightarrow 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 $(0, 0)$.
$x^{2}+y^{2}-4 x-4 y=0$ 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 $
$\Rightarrow -4 x-4 y+16=0$
$\Rightarrow 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 $(0, 0)$.
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