Circles
Circle
star_batch_jee_advanced_2025
Grade None
Question:
Point $M$ moved on the circle $(x-4)^2 + (y-8)^2 = 20$. Then it broke away from it and move along a tangent to the circle, cuts the $x$-axis at the point $(-2, 0)$. The coordinates of a point on the circle at which the moving point broke away is:
-\frac{3}{5}, \frac{46}{5}
-\frac{2}{5}, \frac{44}{5}
6, 4
3, 5
Step-by-Step Solution
Key Concept: Use the angle addition formula with the given sine value to find slopes of both tangents from an external point.
Given $\sin\theta = \frac{2\sqrt{5}}{10} = \frac{1}{\sqrt{5}}$, the slopes of tangents $PA$ and $PB$ are $\tan(\alpha \pm \theta)$. Computing: $\tan\alpha = \frac{8}{3}$, so $\tan(\alpha + \theta) = \frac{11}{5}$ and $\tan(\alpha - \theta) = \frac{5}{10} = \frac{1}{2}$. The chord of contact or tangent pair equations yield points $A$ and $B$ on the circle, with final coordinates confirmed as $(6, 4)$ and related symmetric points.
Correct Answer: 2,3