Question

If the angle between the tangents drawn from P to the circle $x^2 + y^2 + 4x - 6y +9 \sin^2 \alpha + 13 \cos^2 \alpha = 0$ is $2\alpha$, then the locus of P is

• A. x^2 + y^2 + 4x - 6y + 14 = 0
• B. x^2 + y^2 + 4x -6y-4=0
• C. x^2 + y^2 + 4x -6y-9=0
• D. x^2 + y^2 + 4x -6y+9=0

Answer 1

Answer: (D)

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

Answer 2

The given circle equation is $x^2 + y^2 + 4x - 6y + 9 \sin^2 \alpha + 13 \cos^2 \alpha = 0$. The constant term can be rewritten as $c = 9 \sin^2 \alpha + 13 \cos^2 \alpha = 9 + 4 \cos^2 \alpha$. The center of the circle is $(-g, -f) = (-2, 3)$. The radius squared is $r^2 = g^2 + f^2 - c = 4 + 9 - (9 + 4 \cos^2 \alpha) = 4 \sin^2 \alpha$, so the radius is $r = 2|\sin \alpha|$. If P is a point $(h, k)$ and the angle between the tangents from P to the circle is $2\alpha$, then the locus of P is a circle concentric with the given circle. The radius of this locus circle is $R = r / \sin \alpha = (2|\sin \alpha|) / |\sin \alpha| = 2$. The locus is $(h+2)^2 + (k-3)^2 = 2^2 = 4$. Expanding this gives $h^2 + 4h + 4 + k^2 - 6k + 9 = 4$, which simplifies to $h^2 + k^2 + 4h - 6k + 9 = 0$. Therefore, the locus of P is $x^2 + y^2 + 4x - 6y + 9 = 0$.