Question

The image of the circle $x^2+y^2=16$ in the line $x+y=4$ is

• A. $x^2+y^2-4x-4y+16=0$
• B. $x^2+y^2+4x-4y+32=0$
• C. $x^2+y^2-8x-8y+16=0$
• D. $x^2+y^2-8x-8y+32=0$

Answer 1

Answer: (C)

$x^2+y^2-8x-8y+16=0$

The parametric equation of the circle $x^2+y^2=16$ is $x=4\cos \theta ,y=4\sin \theta ,\theta$ being parameter so any point on the circle

$x^2+y^2=16$ is $(4\cos \theta ,4\sin \theta )$. Let $(X,Y)$ be the image of the point $(4\cos \theta ,4\sin \theta )$ in the line $x+y=4$ then,

$\frac{X-4\cos \theta }{1}=\frac{Y-4\sin \theta }{1}=\frac{-2(4\cos \theta +4\sin \theta -4)}{(1^2+1^2)}$

$\Rightarrow X-4\cos \theta =Y-4\sin \theta =-(4\cos \theta +4\sin \theta -4)$

$\Rightarrow X-4\cos \theta =-4\cos \theta -4\sin \theta +4andY-4\sin \theta =-4\cos \theta -4\sin \theta +4$

$\Rightarrow X-4=-4\sin \theta$ ....(i)

$\Rightarrow Y-4=-4\cos \theta$ ....(ii)

$\Rightarrow$ From (i) & (ii) we get

${(X-4)}^2+{(Y-4)}^2=16$

$X^2+Y^2-8X-8Y+16=0$
Hence choice (c) is correct answer..