If a line y = mx + c , is a tangent to the circle (x - 3) ^ 2 + y ^...

Slope of tangent to ${x}^{2} + {y}^{2} = 1$ at \left(\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right)$<br />${x}^{2} + {y}^{2} = 1$<br />$2x + 2y{y}^{'} = 0$<br />${y}^{'} = - \dfrac{x}{y} = - 1$<br />Given that,$y = mx + c$, is perpendicular to${L}_{1}$,<br />So,$m = 1 ⇒ y = x + c$</p><p>Now distance of$\left(3, 0\right)$from$y = x + c$is<br />\left|\dfrac{c + 3}{\sqrt{2}}\right| = 1$
${c}^{2} + 6c + 9 = 2$
${c}^{2} + 6c + 7 = 0$.