Circles
Circle
Allen Star Batch
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: Tangency conditions combined with algebraic constraint equations lead directly to relationships between coordinates without explicit angle elimination.
For part (C), given tangent $x\cos\theta + y\sin\theta = 1$ passes through $(x_1, -1)$ and $(x_2, 1)$, we use the constraint equations $x_1\cos\theta = 1 + \sin\theta$ and $x_2\cos\theta = 1 - \sin\theta$. Multiplying these equations gives $x_1x_2\cos^2\theta = (1+\sin\theta)(1-\sin\theta) = \cos^2\theta$, therefore $x_1x_2 = 1$. For part (D), substituting $t = \frac{1}{r}$ into the circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$ yields $t^3 + 2gt^2 + ct^2 + 2ft + 1 = 0$, and since $a, b, c, d$ are roots, their product $abcd = 1$.
Correct Answer: [A-r] [B-p][C-q] [D-q]