Question

A circle touches a straight line $lx+my+n=0$ and cuts the circle $x^2+y^2=9$ orthogonally. The locus of centres of such circles is-

• A. ${(lx+my+n)}^2=(l^2+m^2)(x^2+y^2-9)$
• B. ${(lx+my-n)}^2=(l^2+m^2)(x^2+y^2-9)$
• C. ${(lx+my+n)}^2=(l^2+m^2)(x^2+y^2+9)$
• D. None of these

Answer 1

The correct option is A ${(lx+my+n)}^2=(l^2+m^2)(x^2+y^2-9)$

Let the equation of the circle is-

$x^2+y^2+2gx+2fy+c=\mathrm{0.......}\left(i\right)$

which touches the line $lx+my+n=0$

$\therefore$$\left|\frac{-lg-mf+n}{\sqrt{l^2+m^2}}\right|=\sqrt{g^2+f^2-c}......\left(ii\right)$

and circle $\left(i\right)$ is orthogonal to the circle $x^2+y^2=9$

$\therefore$ $0\times g+0\times f=c-9$

$\Rightarrow$ $c=9$

from $\left(2\right)$ and $\left(3\right)$

$\left|\frac{-lg-mf+n}{\sqrt{l^2+m^2}}\right|=\sqrt{g^2+f^2-9}$

$\therefore$ locus of $(-g,-f)$ is

${(lx+my+n)}^2=(x^2+y^2-9)(l^2+m^2)$