$A$ circle $C$ of radius $2$ units rolls inside the ring of the circle $x^{2} + y^{2} + 8 x - 2 y - 19 = 0$ . Then the locus of the centre of $C$

x^{2} + y^{2} + 8 x - 2 y - 47 = 0

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x^{2} + y^{2} + 8 x - 2 y - 1 = 0

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x^{2} + y^{2} + 8 x - 2 y + 1 = 0

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x^{2} + y^{2} - 8 x + 2 y + 1 = 0

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Solution

The correct option is B $x^{2} + y^{2} + 8 x - 2 y + 1 = 0$

$S_{1} : x^{2} + y^{2} + 8 x - 2 y - 19 = 0$
$C_{1} = (- 4, 1)$ and $r_{1} = 6$
Let P(x,y) is a centre of the circle C.
Distance $P C_{1} = 6 - 2 = 4$
$\sqrt{(x + 4)^{2} + (y - 1)^{2}} = 4$
$x^{2} + y^{2} + 8 x - 2 y + 17 - 16 = 0$
$\Rightarrow x^{2} + y^{2} + 8 x - 2 y + 1 = 0$

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