Question

The centre of the circle $S=0$ lies on the line $2 x-$ $2 y+9=0$ and $S=0$ cuts orthogonally the circle $x^{2}+y^{2}=4$. Then the circle $S=0$ passes through two fixed points, which lie on

• A. $x=y$
• B. $x+y=0$
• C. $x-2 y=0$
• D. $x+2 y=0$

Answer 1

Answer: (B)

Centre lies on the line $2 x-2 y+9=0$.
So let coordinate of centre be $\left(h, \frac{2 h+9}{2}\right)$.
Let the radius of circle be ' $r$ '.
So equation of circle is
$(x-h)^{2}+\left(y-\frac{2 h+9}{2}\right)^{2}=r^{2}$
$x^{2}+y^{2}-2 h x-y(2 h+9)+2 h^{2}+9 h-r^{2}+\frac{81}{4}=0$
Above circle cuts orthogonally the circle $x^{2}+y^{2}=4$.
so $2 h^{2}+9 h+\frac{65}{4}-r^{2}=0$
or $2 h^{2}+9 h-r^{2}=-\frac{65}{4}$
So equation of required circle is:
$x^{2}+y^{2}-2 h x-y(2 h+9)+4=0$
$\left(x^{2}+y^{2}-9 y+4\right)+h(-2 y-2 x)=0$
So this circle always passes through points of intersection of
$x^{2}+y^{2}-9 y+4=0$ and $x+y=0$.
Therefore fixed points are $(-4,4)$ and $\left(-\frac{1}{2}, \frac{1}{2}\right)$.