The angle made by the tangent to the circle $x^{2} + y^{2} - 8 x + 6 y + 20$ = 0 at (3,-1) with the positive Direction of the x-axis is

$Tan^{-1}\frac{2}{3}$

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Solution

The correct option is D

(3,1)

$x^{2} + y^{2} - 8 n + 6 y + 20$ = 0

Eqn of tgt is $3 x - y - 4 (x + 3) + 3 (y - 1) + 20$ = 0

$\to - n + 2 y + 5$ = 0

$M$ = $\frac{1}{2}$

$T a n θ$ = $\frac{1}{2}$ as $θ$ = $t a n^{- 1} \frac{1}{2}$

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