Mathematics - Circle Question with Solution | TestHub

Slope of tangent to $x^2+y^2=1$ at $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$
$x^2+y^2=1$
$2x+2y{y}^{\text{'}}=0$
$y^{\text{'}}=-\frac{x}{y}=-1$
Given that, $y=mx+c$, is perpendicular to ${\mathrm{L}}_1$,
So, $m=1 \Rightarrow y=x+c$

Now distance of $\left(3,0\right)$ from $y=x+c$ is
$\left|\frac{c+3}{\sqrt{2}}\right|=1$
$c^2+6c+9=2$
$c^2+6c+7=0$.