Question

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

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

Answer 1

Answer: (D)

Let $(h, k)$ be the centre of the circle which touches the
circle $x^2 + y^2 - 6 x - 6y + 14=0$ 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$