Find the value of $h$ if $x^2 + y^2 = h$ 
Consider equation $x^2 + y^2 = h$ that touches the line $y=3x+2$ at some point $P$. Find the value of $h$

I know that $x^2 + y^2 = h$ is a circle with radius $\sqrt{h}$.
Also, since $y = 3x + 2 $ is a tangent, we know that the slope of the radius perpendicular to the tangent is $M_{OP}= -\frac{1}{3}$. I'm not sure how I can determine the value of $h$ though?

The slope of the perpendicular radius is $-\dfrac 13$ and it passes through the center of the circle, $(0,0)$. So its equation is
$y-0 = -\dfrac 13(x - 0)$, which simplifies to $y = -\dfrac 13x$. Avoiding fractions, we write this as $x = -3y$.
We compute the intersection
\begin{align}
   x &= -3y \\
   y &= 3x + 2 \\
\hline
   y &= -9y + 2\\
   y &= \dfrac 15 \\
   x &= -\dfrac 35
\end{align}
So $h = x^2 + y^2 = \dfrac{10}{25} = \dfrac 25$
