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If a line $y = m x + c$ is a tangent to the circle ${(x - 3)}^{2} + y^{2} = 1$ and it is perpendicular to a line $L_{1}$ . Where $L_{1}$ is the tangent to the circle $x^{2} + y^{2} = 1$ at the point $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ then

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$c^{2} + 7 c + 6 = 0$

$c^{2} + 6 c + 7 = 0$

$c^{2} - 6 c + 7 = 0$

$c^{2} - 7 c + 6 = 0$

answer is B.

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Detailed Solution

$x^{2} + y^{2} = 1$

$2 x + 2 y y_{1} = 0$

$y_{1} = - \frac{x}{y}$

Slope $L_{1} :$ $m = (\frac{d y}{d x}) = - 1$

Perpendicular $y = 1 (x) + C$

$\Rightarrow x - y + c = 0$

$∴ r = d$

$\Rightarrow 1 =$ perpendicular distance from A (3,0) To $x - y + c = 0$

$= \frac{|3 - 0 + c|}{\sqrt{1 + 1}}$

$= \sqrt{2} = |c + 3|$

$= c^{2} + 6 C + 9 = 2$

$= c^{2} + 6 C + 7 = 0$

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