The angle between the pair of tangents from a | vedclass

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(D) The circle is $x^2 + y^2 + 4x - 6y + 9\sin^2\alpha + 13\cos^2\alpha = 0$. Its center is $C(-2, 3)$. The radius $r$ is given by $\sqrt{(-2)^2 + (3)

Vedclass · Accepted Answer

$x^2 + y^2 + 4x - 6y + 9 = 0$

Vedclass · Answer

(D) The circle is $x^2 + y^2 + 4x - 6y + 9\sin^2\alpha + 13\cos^2\alpha = 0$. Its center is $C(-2, 3)$. The radius $r$ is given by $\sqrt{(-2)^2 + (3)^2 - (9\sin^2\alpha + 13\cos^2\alpha)} = \sqrt{4 + 9 - 9\sin^2\alpha - 13\cos^2\alpha} = \sqrt{13 - 9\sin^2\alpha - 13\cos^2\alpha} = \sqrt{13(1 - \cos^2\alpha) - 9\sin^2\alpha} = \sqrt{13\sin^2\alpha - 9\sin^2\alpha} = \sqrt{4\sin^2\alpha} = 2\sin\alpha$. Let $P(h, k)$ be the point. The angle between the tangents is $2\alpha$,so the angle between the line $PC$ and a tangent is $\alpha$. In the right-angled triangle $	riangle PAC$,$\sin\alpha = \frac{AC}{PC} = \frac{r}{PC}$. Thus,$\sin\alpha = \frac{2\sin\alpha}{\sqrt{(h+2)^2 + (k-3)^2}}$. This implies $\sqrt{(h+2)^2 + (k-3)^2} = 2$. Squaring both sides,$(h+2)^2 + (k-3)^2 = 4$. Expanding this,$h^2 + 4h + 4 + k^2 - 6k + 9 = 4$. $h^2 + k^2 + 4h - 6k + 9 = 0$. Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 + 4x - 6y + 9 = 0$.

The angle between the pair of tangents from a point $P$ to the circle $x^2 + y^2 + 4x - 6y + 9\sin^2\alpha + 13\cos^2\alpha = 0$ is $2\alpha$. The equation of the locus of $P$ is...

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