If theta is the angle between the tangents dr | vedclass

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(A) The equation of the circle is $x^2 + y^2 - 4x - 6y + c = 0$. The center is $C(2, 3)$ and the radius is $r = \sqrt{2^2 + 3^2 - c} = \sqrt{13 - c}$.

Vedclass · Accepted Answer

$4$

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(A) The equation of the circle is $x^2 + y^2 - 4x - 6y + c = 0$. The center is $C(2, 3)$ and the radius is $r = \sqrt{2^2 + 3^2 - c} = \sqrt{13 - c}$.Let $P = (-1, -1)$. The distance $d$ from $P$ to $C$ is $d = \sqrt{(2 - (-1))^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = 5$.Let $\alpha$ be the angle between the tangent and the line joining the center to the external point. Then $\sin \alpha = \frac{r}{d}$.The angle between the two tangents is $	heta = 2\alpha$,so $\alpha = \frac{	heta}{2}$.Given $\cos 	heta = -\frac{7}{25}$,we use the identity $\cos 	heta = 1 - 2\sin^2 \alpha$.$-\frac{7}{25} = 1 - 2\sin^2 \alpha \implies 2\sin^2 \alpha = 1 + \frac{7}{25} = \frac{32}{25} \implies \sin^2 \alpha = \frac{16}{25}$.Thus,$\sin \alpha = \frac{4}{5}$.Since $\sin \alpha = \frac{r}{d}$,we have $\frac{r}{5} = \frac{4}{5}$,which gives $r = 4$.

If $\theta$ is the angle between the tangents drawn from the point $P(-1, -1)$ to the circle $x^2 + y^2 - 4x - 6y + c = 0$ and $\cos \theta = -\frac{7}{25}$,then the radius of the circle is

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