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

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(D) The equation of a family of circles touching the line $L: x - y = 0$ at the point $(1, 1)$ is given by $(x - 1)^2 + (y - 1)^2 + \lambda(x - y) = 0

Vedclass · Accepted Answer

$2\sqrt{2}$

Vedclass · Answer

(D) The equation of a family of circles touching the line $L: x - y = 0$ at the point $(1, 1)$ is given by $(x - 1)^2 + (y - 1)^2 + \lambda(x - y) = 0$.Since the circle passes through the point $(1, -3)$,we substitute these coordinates into the equation:$(1 - 1)^2 + (-3 - 1)^2 + \lambda(1 - (-3)) = 0$$0 + (-4)^2 + \lambda(4) = 0$$16 + 4\lambda = 0 \Rightarrow \lambda = -4$.Substituting $\lambda = -4$ back into the equation:$(x - 1)^2 + (y - 1)^2 - 4(x - y) = 0$$x^2 - 2x + 1 + y^2 - 2y + 1 - 4x + 4y = 0$$x^2 + y^2 - 6x + 2y + 2 = 0$.The center of the circle is $(-g, -f) = (3, -1)$ and the radius $r$ is given by $\sqrt{g^2 + f^2 - c} = \sqrt{3^2 + (-1)^2 - 2} = \sqrt{9 + 1 - 2} = \sqrt{8} = 2\sqrt{2}$.

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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