The locus of the centre of a circle which touch the circles x2+y2−6x&#x

Discussion Forum : Circles

Question -

The locus of the centre of a circle which touch the circles $x^{2} + y^{2} - 6 x - 6 y + 14 = 0$ externally and also the Y - axis is given by

Options:

A .

x^{2} - 6 y - 7 y + 14 = 0

B .

x^{2} - 10 x - 6 y + 14 = 0

C .

y^{2} - 6 x - 10 y + 14 = 0

D .

y^{2} - 10 x - 6 y + 14 = 0

Answer: Option D
:
D

Let (x, y) be centre of circle which touch y - axis and given circle.
At any point, the radius of this circle will be equal to 'x' units. Since this circle is touching the given circle which has a radius of two units,
$∴$ Distance between centres of the two circles = 2 + x

$\Rightarrow \sqrt{(x - 3)^{2} + (y - 3)^{2}} = 2 + x$

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

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