Find a circle orthogonal to two other circles Given two circles
$x^2+(y-4)^2=4^2$
and 
$(x-8)^2+(y+2)^2=2^2$
I need to find the circle that's orthogonal to both the above circles, and contains the point $P=(8,4).$
Is there an algebraic way to find this circle? Or is a geometric way easier?
Any advice will be greatly appreciated.
Thanks in advance!

Consider $\bigcirc A$ with radius $a$ and $\bigcirc B$ of radius $b$, and $\bigcirc R$ of radius $r$ orthogonal to both.
That $\bigcirc R$ is orthogonal to $\bigcirc A$ means that either point of intersection determines a right triangle with legs $r$ and $a$ and hypotenuse $|AR|$. Likewise for $\bigcirc B$. Thus,
$$a^2 + r^2 = |AR|^2 \qquad\qquad b^2 + r^2 = |BR|^2$$
If point $P$ is on $\bigcirc R$, then also
$$r^2 = |PR|^2$$
Writing $R(h,k)$, the above equations make a non-linear system in unknowns $h$, $k$, $r$. The non-linearness is daunting, but it can be overcome. In fact, the specific nature of your problem makes it easier than it would be in general.
In your problem, we have $A = (0, 4)$, $B = (8,-2)$, $P = (8,4)$, and $a = 4$, $b = 2$. The above equations become
$$\begin{align}
16 + r^2 &= ( h- 0 )^2 + ( k - 4 )^2 \\
4 + r^2 &= ( h - 8 )^2 + ( k + 2 )^2 \\
r^2 &= ( h - 8 )^2 + ( k - 4 )^2
\end{align}$$
where I won't multiply-out the terms. Instead, I'll use the last equation to replace $r^2$ in the first two and get some convenient cancellation (which is what makes your specific problem easier):
$$\begin{align}
16 + ( h - 8 )^2 + ( k - 4 )^2 &= ( h- 0 )^2 + ( k - 4 )^2 \quad \to \quad 16 + ( h - 8 )^2 = h^2 \\
4 + ( h - 8 )^2 + ( k - 4 )^2 &= ( h - 8 )^2 + ( k + 2 )^2 \quad \to \quad \phantom{1}4 + ( k - 4 )^2 = ( k + 2 )^2\\
\end{align}$$
Now, multiplying-out, we see that the quadratic terms $h^2$ and $k^2$ cancel in their respective equations, and we solve to get
$$h = 5 \qquad k = \frac{4}{3} \qquad \text{and, thus} \qquad r^2 = (5-8)^2 + \left(\frac{4}{3} - 4 \right)^2 = \frac{145}{9}$$
so that the equation of $\bigcirc R$ is
$$\left( x - 5 \right)^2 + \left( y - \frac{4}{3} \right)^2 = \frac{145}{9}$$
This agrees with @Apurv's answer.
(If we hadn't gotten the convenient cancellation, we would still have been able to cancel $h^2$ and $k^2$ from the equation pair. This would leave a linear system in $h$ and $k$ that could be solved.)

Note: One can prove @Apurv's orthogonality condition by observing that
$$x^2 + y^2 + 2 m x + 2 n y + c = 0 \quad\text{and}\quad x^2 + y^2 + 2 p x + 2 q y + d = 0$$
represent circles about $U(-m,-n)$ and $V(-p,-q)$, with respective radii $u$ and $v$ satisfying $u^2 = m^2+n^2-c$ and $v^2 = p^2+q^2-d$. Thus,
$$\begin{align}
u^2 + v^2 = |UV|^2 \quad&\implies\quad m^2 + n^2 + p^2 + q^2 - (c + d) = (m-p)^2 + (n-q)^2 \\
&\implies\quad c + d = 2 m p + 2 n q
\end{align}$$
