Circle
Angle Between Tangents
IIT-JEE 1996
Grade 11
Question:
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 $2\alpha$. Then $P$ lies on the circle:
(1) $x^2+y^2+4x-6y+4 = 0$
(2) $x^2+y^2+4x-6y-9 = 0$
(3) $x^2+y^2+4x-6y-4 = 0$
(4) $x^2 + y^2 + 4x - 6y + 9 = 0$
Step-by-Step Solution
Key Concept: The circle has centre $(-2, 3)$ and radius $r = 2 \sin \alpha$. If the angle between tangents from $P$ is $2\alpha$, then $\sin \alpha = r/d$ where $d = |PC|$, giving $d = 2$. The locus $(x + 2)^2 + (y - 3)^2 = 4$ simplifies to option (4).
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (4)