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

Equation of tangent at (3, -1) is
x - 2y - 5 = 0
Slope tan $θ = \frac{1}{2} θ = t a n^{- 1} (\frac{1}{2})$

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2 kg

_{1} = - 2^i - 2^j m/s

_{2} = (^i -^j) m/s

y = [| s i n x | + | c o s x |] a n d x^{2} + y^{2} = 5

The angle between the lines represented by the equation $4 x^{2} - 24 x y + 11 y^{2} = 0$ are

t a n^{- 1} (1 / 2)

\to r = 3 t^i + \sqrt{3} t^{2}^j - 4^k

t

\to r

\to v

\to v

t = \sqrt{3} s .

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