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Byju's Answer

Standard X

Mathematics

Chord Properties of Circles

The tangent t...

Question

The tangent to the circle $C_{1} : x^{2} + y^{2} - 2 x - 1 = 0$ at the point $(2, 1)$ cuts off a chord of length $4$ from a circle $C_{2}$ whose centre is $(3, - 2)$ . The radius of $C_{2}$ is

\sqrt{6}

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\sqrt{2}

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Solution

The correct option is A $\sqrt{6}$
Equation of tangent on $C_{1}$ at $(2, 1)$ is:

$2 x + y - (x + 2) - 1 = 0$

$x + y = 3$

If it cuts off the chord of the circle $C_{2}$ then the equation of the chord is: $x + y = 3$
Distance of the chord from $(3, - 2)$

$d = ∣ ∣ \frac{3 - 2 - 3}{\sqrt{2}} ∣ ∣ = \sqrt{2}$
Length of the chord is $l = 4$

$r^{2} = \frac{l^{2}}{4 + d^{2}}$ where $r$ is the radius of the circle.

$r^{2} = 4 + 2 = 6 \Rightarrow r = \sqrt{6}$

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The tangent to the circle C1:x2+y2−2x−1=0 at the point (2,1) cuts off a chord of length 4 from a circle C2 whose centre is (3,−2). The radius of C2 is

The tangent to the circle $C_{1} : x^{2} + y^{2} - 2 x - 1 = 0$ at the point $(2, 1)$ cuts off a chord of length $4$ from a circle $C_{2}$ whose centre is $(3, - 2)$ . The radius of $C_{2}$ is