The locus of a point such that the tangents drawn from it to the circle $x^{2} + y^{2} - 6x - 8y = 0$ are perpendicular to each other is

x^{2} + y^{2} - 6 x - 8 y - 25 = 0

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x^{2} + y^{2} + 6 x - 8 y - 25 = 0

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x^{2} + y^{2} - 6 x - 8 y + 25 = 0

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None of these

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Solution

The correct option is A $x^{2} + y^{2} - 6 x - 8 y - 25 = 0$
Given circle is
${(x - 3)}^{2} + {(y - 4)}^{2} = 25$
Since locus of point of intersection of two perpendicular tangents is director circle, then it equation is
${(x - 3)}^{2} + {(y - 4)}^{2} = 50 x^{2} + y^{2} - 6 x - 8 y - 25 = 0$

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x^{2} + y^{2} - 6 x - 8 y - 24 = 0

x^{2} + y^{2} - 6 x - 8 y = 0

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x^{2} + y^{2} - 6 x - 8 y - 24 = 0

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