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

Answer 1

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

The center of the given circle is $(3,3)$

Its radius $=\sqrt{3^2+3^2-14}=2$

Let the center of other circle is $(h,k)$

The distance between center of both circle $=$ Sum of radii

And radius of other circle will be $h$ because it is touching the $y$ axis as shown in figure.

So $\sqrt{{(3-h)}^2+(3-k{)}^2}=h+2$

Squaring both side, we get

${(3-h)}^2+(3-k{)}^2={(h+2)}^2$

Solving above equation we get, $k^2-6k-10h+14=0$

Replacing $h$ with $x$ and $k$ with $y$, we get

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