The line x = y touches a circle at the point | vedclass

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Question

Equation of circle is given as $S + \lambda L = 0$

${\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} + \lambda \left( {x - y} \right) =

Vedclass · Accepted Answer

$2\sqrt 2$

Vedclass · Answer

Equation of circle is given as $S + \lambda L = 0$

${\left( {x - 1} ight)^2} + {\left( {y - 1} ight)^2} + \lambda \left( {x - y} ight) = 0$

through $\left( {1, - 3} ight)$

$16 + \lambda  	imes 4 = 0 \Rightarrow \lambda  =  - 4$

$	herefore {\left( {x - 1} ight)^2} + {\left( {y - 1} ight)^2} + 4\left( {x - y} ight) = 0$

$r = 2\sqrt 2 $

The line $x = y$ touches a circle at the point $(1, 1)$. If the circle also passes through the point $(1, -3)$, then its radius is

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