Question

The locus of the centre of the circle which touches externally the circle $x^2+y^2-6x-6y+14=0$ and also touches the $y$ -axis is given by the equations

• A. $x^2-6x-10y+14=0$
• B. $x^2-10x-6y+14=0$
• C. $y^2-6x-10y+14=0$
• D. $y^2-10x-6y+14=0$

Answer 1

The correct option is D $y^2-10x-6y+14=0$

Let the equation of the circle with centre $(g,f)$ be

$x^2+y^2+2gx+2fy+c=0$ .........$\left(1\right)$

Since, circle $\left(1\right)$ touches $y-$axis

$\therefore$radius of circle $=g$

$\Rightarrow \sqrt{g^2+f^2-c}=g$

$\Rightarrow g^2+f^2-c=g^2$

$\Rightarrow c=f^2$

Also,circle $\left(1\right)$ touches externally the given circle

$x^2+y^2-6x-6y+14=0$ then

${(g-3)}^2+{(f-3)}^2={[\sqrt{9+9-14}+\sqrt{g^2+f^2-c}]}^2$

$\Rightarrow g^2-6g+9+f^2-6f+9={[\sqrt{4}+\sqrt{g^2+c-c}]}^2$ since $c=f^2$

$\Rightarrow g^2-6g+9+f^2-6f+9={(2+g)}^2$

$\Rightarrow g^2-6g+9+f^2-6f+9=4+g^2+4g$

$\Rightarrow -10g+f^2-6f+18=4$

$\Rightarrow f^2-10g-6f+18=4$

Replace $g\to x$ and $f\to y$ we get

$y^2-10x-6y+18-4=0$

Hence, the required locus of centre $(g,f)$ is

$y^2-10x-6y+14=0$