Question

Let $x-4=0$ be the radical axis of two circles which are intersecting orthogonally. If $x^2+y^2=36$ is one of those circles, then the other circle is

• A. $x^2+y^2-16x+36=0$
• B. $x^2+y^2-18x+36=0$
• C. $x^2+y^2-18x+24=0$
• D. $x^2+y^2-6x+8y+36=0$

Answer 1

Answer: (B)

We have, equation of one circle
$x^2+y^2=36$
$\Rightarrow x^2+y^2-36=0\dots (i)$
and radical axis of two circle is $x-4=0$
So, equation of other circle is
$x^2+y^2-36+k(x-4)=0$
$x^2+y^2+kx-4k-36=0\dots (ii)$
Both circles are intersecting orthogonally, then
$-4k-36-36=0$
$-4k=72$
$k=-18$
So, equation of required circle
$x^2+y^2-18x-4(-18)-36=0$
$\Rightarrow x^2+y^2-18x+72-36=0$
$\Rightarrow x^2+y^2-18x+36=0$