The angle between the tangents drawn from a p | vedclass

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

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$x^2 + y^2 + 4x - 2y - 31 = 0$

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(A) The given circle is $x^2 + y^2 + 4x - 2y - 4 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 2, f = -1, c = -4$. The center is $C(-g, -f) = (-2, 1)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 1 - (-4)} = \sqrt{9} = 3$. Let $P(h, k)$ be a point on the locus. The angle between the tangents is $60^{\circ}$,so the angle between the line joining the center to $P$ and the radius is $30^{\circ}$. In the right-angled triangle formed by the center,the point of contact,and $P$,we have $\sin(30^{\circ}) = \frac{r}{CP}$. $\frac{1}{2} = \frac{3}{CP} \implies CP = 6$. $CP^2 = 36$. $(h + 2)^2 + (k - 1)^2 = 36$. $h^2 + 4h + 4 + k^2 - 2k + 1 = 36$. $h^2 + k^2 + 4h - 2k - 31 = 0$. Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 + 4x - 2y - 31 = 0$.

The angle between the tangents drawn from a point $P$ to the circle $x^2 + y^2 + 4x - 2y - 4 = 0$ is $60^{\circ}$. The locus of $P$ is:

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