Question

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

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

Answer 1

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

Let the center of G is $\alpha$ and $\beta$

The equation of circle $C_2$ is $x^2+y^2-6x-6y+14=0$

$\Rightarrow 2g=-6\Rightarrow g=-3$

$\Rightarrow 2f=-6\Rightarrow f=-3$

Centre of $C_2=(-g,-f)=(3,3)$

Radius of $C_2=\sqrt{g^2+f^2-c}=\sqrt{9+9-14}=\sqrt{4}=2$

Now, By distance formula,

$\Rightarrow (3-\alpha {)}^2+(3-\beta {)}^2=(2+\alpha {)}^2$

$\Rightarrow g+{\alpha }^2-6\alpha +99+{\beta }^2-6\beta =4+{\alpha }^2+4\alpha$

${\beta }^2-6\beta -10\alpha +14=0$

Now, replacing $\beta =y$ and $\alpha =x$ we get

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