The locus of a point which moves such that tangent from it to the two circles $x^{2} + y^{2} - 5 x - 3 = 0$ and $3 x^{2} + 3 y^{2} + 3 x + 4 y - 6 = 0$ are equal, is

2 x^{2} + 2 y^{2} + 7 x + 4 y - 3 = 0

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17 x + 4 y + 3 = 0

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4 x^{2} + 4 y^{2} - 3 x + 4 y - 9 = 0

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13 x - 4 y + 15 = 0

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Solution

The correct option is A $17 x + 4 y + 3 = 0$
Given, ${3 x}^{2} + 3 y^{2} + 2 x + 4 y - 6 = 0$
$\Rightarrow x^{2} + y^{2} + (\frac{2}{3}) x + (\frac{4}{3}) y - 2 = 0$ ...(1)
and $x^{2} + y^{2} - 5 x - 3 = 0$ ...(2)
solving (1) and (2) we locus, i.e., $x^{2} + y^{2} - 5 x - 3 = x^{2} + y^{2} + (\frac{2}{3}) x + (\frac{4}{3}) y - 2$
$\Rightarrow 17 x + 4 y + 3 = 0.$

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x^{2} + y^{2} - 5 x - 3 = 0

3 x^{2} + 3 y^{2} + 2 x + 4 y - 6 = 0

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