The locus of a point from which the length of tangents to the circles $x^{2} + y^{2} = 4$ and $2 (x^{2} + y^{2}) - 10 x + 3 y - 2 = 0$ are equal, is

a straight line inclined at with the line joining the centres of the circles

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a circle

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an ellipse

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a straight line perpendicular to the line joining the centres of the circles

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Solution

The correct option is B a circle
Let the point $P (x_{1}, y_{1})$ from which length of tangent to both the circles are equal
Length of tangent $= \sqrt{S_{11}}$
$L_{1} = \sqrt{{x_{1}}^{2} + {y_{1}}^{2} - 4} L_{2} = \sqrt{2 ({x_{1}}^{2} + {y_{1}}^{2}) - 10 x_{1} + 3 y_{1} - 2}$
Given, $L_{1} = L_{2}$
$\sqrt{{x_{1}}^{2} + {y_{1}}^{2} - 4} = \sqrt{2 ({x_{1}}^{2} + {y_{1}}^{2}) - 10 x_{1} + 3 y_{1} - 2} {x_{1}}^{2} + {y_{1}}^{2} - 4 = 2 ({x_{1}}^{2} + {y_{1}}^{2}) - 10 x_{1} + 3 y_{1} - 2 0 = {x_{1}}^{2} + {y_{1}}^{2} - 10 x_{1} + 3 y_{1} + 2$
This is a equation of circle with centre $(5, \frac{- 3}{2})$ and radius $= \frac{\sqrt{101}}{2} u n i t s$

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