<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] So we have equation $x^2 + y^2 = r^2$, who’s geometrical representation would be a circle with the radius $r$. Now we also have an equation of a line represented by: $y = mx + c$ that touches the circle at point $P$. To find this point we will use the method of solving simultaneous equation where one is a quadratic and the other is linear so we will take the value of y and replace it in the first equation as follows:
$$x^2 + (3x + 2)^2 = h$$
$$x^2 + 9x^2 + 2*3x*2 + 4 = h$$
$$10x^2 + 12x + 4 = h$$
$$10x^2 + 12x + 4 - h = 0$$
so now we will have something like:
$$x^2 + (mx + c)^2 = r^2$$ 
$$x^2 + m^2x^2 + 2mcx + c^2 = r^2$$ 
$$(1 + m^2)x^2 + 2mcx + c^2 - r^2 = 0$$
since this equation is a quadratic equation in x, something like $(ax^2 + bx + c = 0)$ where we have:
$a = 10$; $b = 12$; $c = 4 - h$;
using the values in discriminant $(b^2 - 4ac)$: 
$$b^2 - 4ac = (2mc)^2 - 4(1 + m^2)(c^2 - r^2)$$
              $$= 4m^2c^2 - 4(c^2 + m^2c^2 - r^2 - m^2r^2)$$
                $$ = 4m^2c^2 - 4c^2 - 4m^2c^2 + 4r^2 + 4m^2r^2$$
                $$ = -4c^2 + 4r^2 + 4m^2r^2$$
if the discriminant is zero, our equation will have equal roots and the line intersects the circle in one single point which is our case also. So because this line is a tangent to the circle we know:
$$-4c^2 + 4r^2 + 4m^2r^2 = 0$$
$$4r^2 + 4m^2r^2 = 4c^2$$
$$c^2 = r^2 + m^2r^2$$
this will be the condition for a line to be tangent to the circle, so replacing with our values we will have the tangent:
$$(4 - h)^2 = h + 3^2h$$
$$h^2 - 8h + 16 = h + 9h$$
$$h^2 - 18h = -16$$
$$h(h - 18) = -16$$
since: 
$$(h-9)^2 = (h-9)(h-9)$$
$$h^2 - 18h + 81$$ 
$$h(h -18) + 81$$ 
we can write:
$$(h-9)^2 - 65 = 0$$ so:
$$h = 9 \pm \sqrt65$$
I hope this helps
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