<s>[INST] 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?
 [/INST] You're almost there! We want to find the equation of line $OP$. We know its slope, and (since it passes through the origin) we know its $y$-intercept, so its equation is:
$$
y = \tfrac{-1}{3}x
$$
We can now find the intersection point of the radius and tangent by solving the system of equations. Equating, we obtain:
\begin{align*}
\tfrac{-1}{3}x = y = 3x + 2 
&\implies -x = 9x + 6 \\
&\implies -10x = 6 \\
&\implies x = \tfrac{-3}{5} \\
&\implies y = \tfrac{-1}{3} \cdot \tfrac{-3}{5} = \tfrac{1}{5}
\end{align*}
Thus, we conclude that:
$$
h = x^2 + y^2 = (\tfrac{-3}{5})^2 + (\tfrac{1}{5})^2 = \tfrac{9 + 1}{25} = \boxed{\tfrac{2}{5}}
$$
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