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 $(h,k)$ be the centre of the circle which touches the
circle $x^2+y^2-6x-6y+14=0$ and $X$-axis.
The centre of given circle is $(3,3)$ and radius is
$\sqrt{3^2+3^2-14}=\sqrt{9+9-14}=2$
Since, the circle touches y-axis, the distance from its centre to $y$-axis must be equal to its radius, therefore its radius is $h$. Again, the circles meet externally, therefore
the distance between two centres = sum of the radii of the two circles.
Hence, $(h-3{)}^2+(k-3{)}^2=(2+h{)}^2$
$h^2+9-6h+k^2+9-6k=4+h^2+4h$
i.e.$k^2-10h-6h+14=0$
Thus, the locus of $(h,k)$ is
$y^2-10x-6y+14=0$