Find the value of ' $c$ ' if the line $y = c$ is tangent to the circle $x^{2} + y^{2} - 2 x + 2 y - 2 = 0$ at the point $(1, 1)$

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Solution

Equation of circle,
$x^{2} + y^{2} - 2 x + 2 y - 2 = 0... (i)$
General equation of circle,
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0.... (i i)$
Comparing equation (i) and (ii),
$2 g = - 2 \Rightarrow g = - 1$
$2 f = 2 f = 1$
$c = - 2$
Centre of circle $= (- g, - f) = (1, - 1)$
Radius of circle $= \sqrt{g^{2} + f^{2} - c}$
$= \sqrt{(- 1)^{2} + (1)^{2} - (- 2)}$
$= \sqrt{1^{2} + 1^{2} + 2} = \sqrt{1 + 1 + 2}$
$= \sqrt{4} = 2$
Line $y = c$ touches circle at point $(1, 1)$
$1 = c \Rightarrow c = 1$
Thus, value of $c$ is $1$ .

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