Question

If the circle passes through the point $(a,b)$ and cuts the circle $x^2+y^2=k^2$ orthogonally, then the locus of its centre is given by:

• A. $2ax+2by-\left(a^2+b^2+k^2\right)=0$
• B. $2ax+2by+\left(a^2+b^2-k^2\right)=0$
• C. $2ax+2by+\left(a^2+b^2+k^2\right)=0$
• D. None of the above

Answer 1

Answer: (A)

If $S_1=x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$
and $S_2=x^2+y^2+2g_{2}x+2f_{2}y+c_2=0$
are two circles orthogonal to each other, if
$2g_1{g}_2+2f_1{f}_2=c_1+c_2$
Let the equation of circle be
$x^2+y^2+2gx+2fy+c=0\dots (i)$
Its centre is $(-g,-f)$.
This is orthogonal to given circle
$x^2+y^2=k^2$
$\Rightarrow g(0)+f(0)=c-k^2$
$\Rightarrow c=k^2$
$\therefore$ From Eq. (i)
$x^2+y^2+2gx+2fy+k^2=0$
Also, this circle passes through $(a,b)$
$\therefore a^2+b^2+2ga+2fb+k^2=0$
Locus of centre $(-g,-f)$ is
$a^2+b^2-2xa-2yb+k^2=0$
$\Rightarrow 2ax+2by-a^2-b^2-k^2=0$