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The tangent to the circle $C_1 : x^2 + y^2 - 2x- 1\, = 0$ 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 $
$2$
$\sqrt 2 $
$3$
Solution
Here, equation of tangent on ${C_1}$ at $\left( {2,1} \right)$ is:
$2x + y – \left( {x + 2} \right) – 1 = 0$
Or $x + y = 3$
If it cuts off the chord of the circle ${C_2}$ then
the equation of the chord is :
$x + y = 3$
$\therefore $ distance of the chord from $(3,-2)$ is:
$d = \left| {\frac{{3 – 2 – 3}}{{\sqrt 2 }}} \right| = \sqrt 2 $
Also, length of the chord is $l=4$
$\therefore $ redius of ${C_2}\,\, = r = \sqrt {{{\left( {\frac{l}{2}} \right)}^2} + {d^2}} $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{\left( 2 \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} = \sqrt 6 $
