G3.3If $c$ is the largest slope of the tangents from the point $A\left(\frac{\sqrt{10}}{2}, \frac{\sqrt{10}}{2}\right)$ to the circle $C: x^{2}+y^{2}=1$, find the value of $c$.
Let the equation of tangent be $y-\frac{\sqrt{10}}{2}=c\left(x-\frac{\sqrt{10}}{2}\right)$
$$
c x-y+\frac{\sqrt{10}}{2}(1-c)=0
$$

Distance form centre $(0,0)$ to the straight line $=$ radius
$$
\begin{array}{l}
\left|\frac{0-0+\frac{\sqrt{10}}{2}(1-c)}{\sqrt{c^{2}+(-1)^{2}}}\right|=1 \\
\frac{5}{2}(1-c)^{2}=c^{2}+1 \\
5-10 c+5 c^{2}=2 c^{2}+2 \\
3 c^{2}-10 c+3=0 \\
(3 c-1)(c-3)=0
\end{array}
$$
$c=\frac{1}{3}$ or 3. The largest slope $=3$.