JEE PYQ: Circle Question 20
Question 20 - 2020 (07 Jan Shift 2)
Let the tangents drawn from the origin to the circle, $x^2 + y^2 - 8x - 4y + 16 = 0$ touch it at the points $A$ and $B$. The $(AB)^2$ is equal to:
(1) $\frac{52}{5}$
(2) $\frac{56}{5}$
(3) $\frac{64}{5}$
(4) $\frac{32}{5}$
Show Answer
Answer: (3)
Solution
Center $(4, 2)$, $r = 2$. Length of tangent from origin $= \sqrt{16} = 4$. $OA = 4$, $OC = \sqrt{20}$. Chord of contact: $4x + 2y - 16 = 0$. Distance from $C$ to chord $= \frac{|16+4-16|}{\sqrt{20}} = \frac{4}{\sqrt{20}}$. $AB = 2\sqrt{4 - \frac{16}{20}} = 2\sqrt{\frac{64}{20}}$. $AB^2 = \frac{64}{5}$.
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