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

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(A) The equation of the circle $C_1$ is $x^2 + y^2 - 2x - 1 = 0$.The equation of the tangent at point $(x_1, y_1) = (2, 1)$ is given by $xx_1 + yy_1 -

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$\sqrt{6}$

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(A) The equation of the circle $C_1$ is $x^2 + y^2 - 2x - 1 = 0$.The equation of the tangent at point $(x_1, y_1) = (2, 1)$ is given by $xx_1 + yy_1 - (x + x_1) - 1 = 0$.Substituting the values,we get $2x + y - (x + 2) - 1 = 0$,which simplifies to $x + y - 3 = 0$.This line acts as a chord for the circle $C_2$ with center $(3, -2)$.The perpendicular distance $d$ from the center $(3, -2)$ to the line $x + y - 3 = 0$ is $d = \frac{|3 - 2 - 3|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2}$.The length of the chord is $l = 4$,so half the length of the chord is $\frac{l}{2} = 2$.The radius $r$ of the circle $C_2$ is given by $r = \sqrt{(\frac{l}{2})^2 + d^2}$.Substituting the values,$r = \sqrt{2^2 + (\sqrt{2})^2} = \sqrt{4 + 2} = \sqrt{6}$.

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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