The angle between the pair of tangents drawn | vedclass

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(A) The given equation of the circle is $x^2+y^2+4x+4y-1=0$.Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=2, c=-1$.The center of the circle is $

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$\frac{\pi}{2}$

Vedclass · Answer

(A) The given equation of the circle is $x^2+y^2+4x+4y-1=0$.Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=2, c=-1$.The center of the circle is $O(-g, -f) = (-2, -2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+4-(-1)} = \sqrt{9} = 3$.Let the external point be $C(1, 1)$. The distance $OC$ is $\sqrt{(1 - (-2))^2 + (1 - (-2))^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$.Let $\alpha$ be the angle between the line joining the center to the external point and the tangent. In the right-angled triangle $	riangle OAC$ (where $A$ is the point of tangency),$\sin \alpha = \frac{OA}{OC} = \frac{r}{OC} = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}}$.Thus,$\alpha = 45^\circ$ or $\frac{\pi}{4}$.The angle between the pair of tangents is $2\alpha = 2 	imes \frac{\pi}{4} = \frac{\pi}{2}$.

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

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