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The locus of the centre of a circle whic...

The locus of the centre of a circle which touches externally the circle `x^2 + y^2-6x-6y+14 = 0` and also touches 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`

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`

Text Solution

Verified by Experts

The correct Answer is:
D
Let (h,k) be the centre of the circle which touches the circle `x^(2)+y^(2)-6x-6y+14=0` and Y-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 be Y-axis must be equal to its radius radius, therefore itssss 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-6k+14=0`
Thus, the locus of (h, k) is
`y^(2)-10x-6y+14=0`
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