Tangents drawn from the point $(4, 3)$ to the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ are inclined at an angle

\frac{π}{6}

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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Solution

The correct option is D $\frac{π}{2}$
$(4, 3)$ lies outside the circle, so combined equation of the tangents drawn from $(4, 3)$ to the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ is $S S_{1} = T^{2}$
or $(x^{2} + y^{2} - 2 x - 4 y) (16 + 9 - 8 - 12) = {[4 x + 3 y - (x + 4) - 2 (y + 3)]}^{2}$
$\Rightarrow 4 x^{2} - 4 y^{2} - 50 x + 6 x y + 100 = 0$
since coefficient of $x^{2} +$ coefficient of $y^{2} = 4 + (- 4) = 0$
$∴$ Pair of tangents is perpendicular

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