The angle between the pair of tangents f | vedclass

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Question

$x^{2}+y^{2}+4 x+2 y-4=0$

$1+\frac{1}{4}+x+1-4+20$

Peent ciss (Cutside the circle $m(x-1)=\left(1-\frac{1}{2}\right) \Rightarrow 2 \operatorname

Vedclass · Accepted Answer

$\sin^{-1}\,\frac{4}{5}$

Vedclass · Answer

$x^{2}+y^{2}+4 x+2 y-4=0$

$1+\frac{1}{4}+x+1-4+20$

Peent ciss (Cutside the circle $m(x-1)=\left(1-\frac{1}{2}ight) \Rightarrow 2 \operatorname{sen}(x-1)=(y-1)$

$(2 m x-2 m-1)=y$

$8 m x-2 m-1)^{2}+x^{2}+4 x+2 m x-4 m-2$

$-4=0$

by soluting meget

$m=\frac{3}{4}=	an 	heta$

The angle between the pair of tangents from the point $(1, 1/2)$ to the circle $x^2 + y^2 + 4x + 2y -4=0$ is-

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