Question

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

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

Answer 1

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

$x^2+y^2+6x-6y+14=0$
$(x+3{)}^2+(y-3{)}^2-18+14=0$
$(x+3{)}^2+(y-3{)}^2=4$
Hence, the radius is 2 and centre is (-3, 3).
Let the centre of the required circle be $(x,y)$
Since it touches the y axis, its radius will be $-x$
Since this circle touches the other circle externally, the distance between the centres must be equal to the sum of radii.
Hence,
$\sqrt{(x--3{)}^2+(y-3{)}^2}=-x+2$
On squaring,
$(x+3{)}^2+(y-3{)}^2=(2-x{)}^2$
$x^2+6x+9+y^2-6y+9=4+x^2-4x$
$y^2+10x-6y+14=0$