To continue the way you’ve started, you know that the distance from the circle’s center $C$ to the point $P=(0,12)$ is $\sqrt{85}$ and the radius of the circle is $\sqrt{17}$, so the distance from $P$ to the points of tangency is $\sqrt{85-17}=2\sqrt{17}$. The points of tangency are thus the intersections of the given circle with the circle $x^2+(y-12)^2=68$.
To solve this system of equations, start by expanding them into $$\begin{align} x^2+y^2-12x-10y+44 &= 0 \\ x^2+y^2-24y+76 &= 0 \end{align}$$ and then subtracting one from the other to eliminate the quadratic terms: $$12x-14y+32=0.$$ You can now solve this equation for $x$ or $y$ and then substitute back into one of the circle equations to get a quadratic equation in $x$ or $y$ that I expect you know how to solve. Since the second circle’s equation only has a single terms that involves $x$, I’ll solve for $x$ and substitute into that one: $$x = \frac76y-\frac83\tag1$$ and, after some manipulation, $${17\over36}(5y^2-64y+176)=0.\tag2$$ Solve equation (2) for $y$ and get the corresponding $x$ values from equation (1). You then have the two points of tangency and I hope that you can build the equations of the tangent lines from them.
Here is a different, much less involved way to solve this problem. You can tell from the diagram (and can prove mathematically) that neither tangent will be vertical, so the equations of the tangent lines can be written in the form $y=mx+12$. Substitute this into the equation of the circle and rearrange: $$(x-6)^2+(mx+12-5)^2 = 17 \\ (m^2+1)x^2+(14m-12)x+68 = 0.$$ For the line to be tangent to the circle, this equation must have a repeated real root, i.e., the discriminant $(14m-12)^2-4\cdot68(m^2+1)$ must be equal to zero. Solve this equation for $m$ and you have the equations of the tangent lines without further ado.
