Circles
Circle
star_batch_jee_advanced_2025
Grade 11
Question:
MATCH THE FOLLOWING:
(A) If the line $2x - y + 1 = 0$ is tangent to the circle at the point $(2, 5)$ whose centre lies on the line $x - 2y = 4$, then radius of this circle is
(B) Triangle $ABC$ is right angled at $A$. The circle with centre $A$ and radius $AB$ cuts $BC$ and $AC$ internally at $D$ and $E$ respectively. If $BD = 20$ and $DC = 16$ then the length $AC$ equals
(C) Let $C$ be the circle of radius unity centred at the origin. If two positive numbers $x_1$ and $x_2$ are such that the line passing through $(x_1, -1)$ and $(x_2, 1)$ is tangent to $C$ then $x_1 x_2$ is:
(D) If $\left(a, \frac{1}{a}\right), \left(b, \frac{1}{b}\right), \left(c, \frac{1}{c}\right)$ and $\left(d, \frac{1}{d}\right)$ are four distinct points on a circle of radius $4$ units then, $abcd$ is equal to
Step-by-Step Solution
Key Concept: Use angle-slope relationships and power of a point theorem to connect geometric constraints to circle parameters.
For option (A), $\tan\theta = \frac{2 - \frac{1}{2}}{1 + 2 \cdot \frac{1}{2}} = \frac{3}{4}$. Since $\sqrt{8^2 + 4^2} = 4\sqrt{5}$ and $\tan\theta = \frac{3}{4}$, the radius $r = 3\sqrt{5}$. For option (B), using the power of a point and the chord-intersection property, $(AC-r)(AC+r) = 16 \times 36$, which combined with $(AC)^2 + r^2 = 36^2$ gives $AC = 6\sqrt{26}$.
Correct Answer: [A-r] [B-p][C-q] [D-q]