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(B) The given circles are $S_1: x^2 + y^2 - 5x - 3 = 0$ and $S_2: x^2 + y^2 + \frac{2}{3}x + \frac{4}{3}y - 2 = 0$. Let $P(x, y)$ be a point such that

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$17x + 4y + 3 = 0$

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(B) The given circles are $S_1: x^2 + y^2 - 5x - 3 = 0$ and $S_2: x^2 + y^2 + \frac{2}{3}x + \frac{4}{3}y - 2 = 0$. Let $P(x, y)$ be a point such that the lengths of the tangents from $P$ to both circles are equal. The length of the tangent from $P(x, y)$ to a circle $S = 0$ is $\sqrt{S}$. Thus,$\sqrt{x^2 + y^2 - 5x - 3} = \sqrt{x^2 + y^2 + \frac{2}{3}x + \frac{4}{3}y - 2}$. Squaring both sides,we get $x^2 + y^2 - 5x - 3 = x^2 + y^2 + \frac{2}{3}x + \frac{4}{3}y - 2$. Simplifying,$-5x - 3 = \frac{2}{3}x + \frac{4}{3}y - 2$. Multiplying by $3$,$-15x - 9 = 2x + 4y - 6$. Rearranging the terms,$17x + 4y + 3 = 0$. This is the equation of the radical axis of the two circles.

Find the locus of a point from which the lengths of the tangents drawn to the two circles $x^2 + y^2 - 5x - 3 = 0$ and $3x^2 + 3y^2 + 2x + 4y - 6 = 0$ are equal.

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