Question

The centre of the circle $S=0$ lies on the line $2x-$ $2y+9=0$ 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. $x-2y=0$
• D. $x+2y=0$

Answer 1

Answer: (B)

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