Question

The equation of the system of coaxal circles that are tangent at $(\sqrt{2},4)$ to the locus of the point ofintersection of mutually $\perp$ tangents to the circle $x^2+y^2=9$, is

• A. $(x^2+y^2-18)+\lambda (\sqrt{2}x+4y-18)=0$
• B. $(x^2+y^2-18)+\lambda (4x+\sqrt{2}y-18)=0$
• C. $(x^2+y^2-16)+\lambda (\sqrt{2}x+4y-16)=0$
• D. None of these

Answer 1

Answer: (A)

$(x^2+y^2-18)+\lambda (\sqrt{2}x+4y-18)=0$

Centre of the circle $x^2+y^2=9$ is $(0,0)$ and any tangent to the circle is

$x\cos \alpha +y\sin \alpha =3$ ...(1)

Its distance from centre $(0,0)$ is equal to radius $3.$

Any tangent to $x^2+y^2=9$ but $\perp$ to (1) is obtained by replacing $\alpha$ by $(\alpha -{90}^0)$ and its equation is

$x\cos (\alpha -{90}^0)+y\sin (\alpha -{90}^0)=3$

$\Rightarrow x\cos (\alpha -{90}^0)-y\sin ({90}^0-\alpha )=3$

$\Rightarrow x\sin \alpha -y\cos \alpha =3$ ..(2)

Squaring and adding (1) and (2) we get

$x^2+y^2=18$ which is a circle concentric with the given circle.

$\therefore$ Locus is $S\equiv x^2+y^2-18=0$ ...(3)

Equation of tangent to (3) at $(\sqrt{2},4)$ is

$P\equiv \sqrt{2}x+4y-18=0.$

$\therefore$ System of coaxal circles is $S+\lambda P=0.$