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  4. If a line, y — mx + c is a tangent to the circle, (x — 3)2 + y2 = 1 and it is perpendicular to a line L1, where L1 is the tangent to the circle, x2+y2=1 at the point ((1/√2), (1/√2)) ; then :

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Q. If a line, $y — mx + c$ 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 $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) ;$ then :

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Solution:

Slope of tangent to $x^{2} + y^{2} = 1$ at $\frac{P}{ }\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$2x + 2yy' = 0 \Rightarrow m_{T P} = -1$
$y = mx + c$ is tangent to $\left(x - 3\right)^{2} + y^{2} = 1$
$y = x + c$ is tangent to $\left(x - 3\right)^{2} + y^{2} = 1$
$\left|\frac{c+3}{\sqrt{2}}\right| = 1 \Rightarrow c^{2 }+ 6c + 7 = 0$