The tangent to the circle C_1 : x^2 + y^2 - 2 | vedclass

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Question

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

Vedclass · Accepted Answer

$\sqrt 6 $

Vedclass · Answer

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

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

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