Question

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

• 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

Answer: (D)

Let the centre of the circle be $\left(h,k\right)$. Since the circle touches the axis of $y$.
$\therefore$ Its radius will be $r_1=h$.
Centre of the other given circle is $\left(3,3\right)$ and radius, $r_2=2$.
Since circles touch externally so distance between centres
$=r_1+r_2=h+2$
$\Rightarrow {\left(h-3\right)}^2+{\left(k-3\right)}^2={\left(h+2\right)}^2$
$\Rightarrow k^2-10h-6k+14=0$
$\therefore$ Required locus is $y^2-10x-6y+14=0$.