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

The correct option is D

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

Step 1: Find the radius and centre of the given circle

In the question, a circle $x^2+y^2-6x-6y+14=0$ is given.

Rewrite the equation of the given circle as follows:

$$\begin{gathered} x^2+y^2-6x-6y+14=0 \\ \Rightarrow x^2-6x+9+y^2-6y+5+4=4 \\ \Rightarrow x^2-6x+9+y^2-6y+9=4 \\ \Rightarrow {\left(x-3\right)}^2+{(y-3)}^2=4 \end{gathered}$$

So, the given equation of circle can be rewritten as ${\left(x-3\right)}^2+{\left(y-3\right)}^2=4$.

So, the centre of the given circle is $(3,3)$ and the radius is $2$.

Step 2: Find the equation of the locus
Assume that, $(h,k)$ be the centre of a circle, which touches externally the given circle and also touches the y-axis.

Since the assumed circle touches the y-axis. so, its radius is h.

We know that the condition of circles to touch externally is: $C_1{C}_2=r_1+r_2$ where $C_1{C}_2$ is the distance between the centers of two circles and $r_1,r_2$ be the radii.

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

Taking square on both sides,

$$\begin{gathered} {(h-3)}^2+{(k-3)}^2={\left(h+2\right)}^2 \\ \Rightarrow h^2+9-6h+k^2+9-6k=h^2+4+4h \\ \Rightarrow -10h+k^2-6k+14=0 \\ \Rightarrow k^2-6k-10h+14=0 \end{gathered}$$

Now, replace h with x and k with y.

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

Therefore, 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: $y^2-10x-6y+14=0$.

Hence, the correct answer is option D.