The real intersection between these two equations is about $x=18.39$ and $x=11.81$. However, I keep coming up with imaginary solutions. Please explain how I can solve for those values.
My work so far
$(x-10)^2+(-2|x-10|+9)^2=6^2$
$x^2-20x+100+4|x^2-20x+100|-36|x-10|+81=36$
$4|x^2-20x+100|-36|x-10|=-x^2+20x-64$
Intuitively, the first expression is a circle centered at $(10,10)$ with radius $6$, and the second expression is V-shaped, with the turning point at (10,19), directly above the circle. Therefore, one can quickly see that there are four intersection points.
These absolute values in $y=-2|x-10|+19$ can make things messy, so why not remove them from the very beginning.
When $x\geq10$, $y=-2x+39$.
When $x<10$, $y=2x-1$.
Then substitute the corresponding expressions for $y$ in $\left(x-10\right)^{2}+\left(y-10\right)^{2}=6^{2}$. Then we are only left with two simple quadratic equations, namely:
$5 x^2-136 x+905=0$ (for $x\geq10$)
$5 x^2-64 x+185=0$ (for $x<10$)
which you can solve anyway you like (complete the square, quadratic formula, with a calculator, etc.)
Finally you get: $$x=\frac{1}{5} \left(32-3 \sqrt{11}\right)\land y=\frac{1}{5} \left(59-6 \sqrt{11}\right)$$ $$x=\frac{1}{5} \left(68-3 \sqrt{11}\right)\land y=\frac{1}{5} \left(6 \sqrt{11}+59\right)$$ $$x=\frac{1}{5} \left(3 \sqrt{11}+32\right)\land y=\frac{1}{5} \left(6 \sqrt{11}+59\right)$$ $$x=\frac{1}{5} \left(3 \sqrt{11}+68\right)\land y=\frac{1}{5} \left(59-6 \sqrt{11}\right)$$
(Decimal form answer is shown along with the plot.)
And that's it.