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

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$13 x - 4 y + 15 = 0$

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

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

$17 x + 4 y + 3 = 0$

answer is B.

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Detailed Solution

Let $P (h, k)$ be the coordinates of a point from which equal tangents are drawn to the circles $x^{2} + y^{2} - 5 x - 3 = 0$ and $3 x^{2} + 3 y^{2} + 2 x + 4 y - 6 = 0$ . Then,

$h^{2} + k^{2} - 5 h - 3 = h^{2} + k^{2} + \frac{2}{3} h + \frac{4}{3} k - 2$

$\Rightarrow 17 h + 4 + 3 = 0$

Hence, the locus of $(h, k)$ is $17 x + 4 y + 3 = 0$

$A L I T E R$ Required locus is the radical axis of the given circles and its equation is given by

$\begin{array}{l} S_{1} - S_{2} = 0 \\ \Rightarrow (x^{2} + y^{2} - 5 x - 3) - (x^{2} + y^{2} + \frac{2}{3} x + \frac{4}{3} y - 2) = 0 \\ \Rightarrow 17 x + 4 y + 3 = 0 \end{array}$