Solving these two equations simultaneously I'm having a hard time to solve these two equations simultaneously. I'm arriving to a very long equation..
$$x_0^2+y_0^2=(7\sqrt{2})^2=98$$
$$\sqrt{25+(x_0+2)^2}+\sqrt{4+(y_0-5)^2}=7\sqrt{2}$$
 A: Drop subscripts of $x$ and $y$. 
Put $u=x+2, v=y-5$.
Then the equations become
$$(u-2)^2+(v+5)^2=(7\sqrt{2})^2 \qquad \cdots (1)$$
and
$$\sqrt{25+u^2}+\sqrt{4+v^2}=7\sqrt{2}\qquad \cdots (2)$$
Squaring $(2)$ equals $(1)$, i.e. 
$$\begin{align}(25+u^2)+(4+v^2)+2\sqrt{(25+u^2)(4+v^2)}&=(u^2-4u+4)+(v^2+10v+25)\\
\sqrt{(25+u^2)(4+v^2)}&=-2u+5v\\
\end{align}$$
Squaring:
$$\begin{align}
100+4u^2+25v^2+u^2v^2&=4u^2-20uv+25v^2\\
(uv)^2+20uv+100&=0\\
(uv+10)^2&=0\\
uv&=-10 \Rightarrow v=-\frac{10}u\end{align}$$
Substituting back into $(1)$:
$$\begin{align} 
(u-2)^2+(-\frac {10}u+5)^2&=98\\
\end{align}$$
Solving numerically gives
$$\begin{align}u&=-5, -2.6893, \quad \ \; 0.6752,\  11.0142\\
v=-\frac {10}u &=\quad 2,\ \  3.7184, \; -14.8104, -0.9079\\
x=u-2&=-7, -4.6893, \ -1.3248, \ \ \   9.0142\\
y=v+5&=\ \ 7,\quad 8.7184, \ -9.8104, \ \ \ 4.0921
\end{align}$$
Checking by substitution shows that only the first two sets of numbers are valid, hence solution is 
$$(x,y)=(-7,7), (-4.6893, 8.7184)\qquad \blacksquare$$
A: This is not an answer yet, but perhaps it helps: 
The first equation represents a circle of radius $7\sqrt{2}$ with center $0$. The second equation is obviously a bounded and symmetrical set around its center around $(-2,5)$
I did a quick plot and there might be a solution but it seems to be no more than one.

A: Use polar coordinates $(x,y) = (-r \sin \theta, r \cos \theta)$ to get $r=\sqrt{98}$ and
$$ \sqrt{98 \cos^2 \theta - 2 \sqrt{98} (5) \cos \theta + (-2)^2 + (5)^2} + \\
\sqrt{98 \sin^2 \theta+2 \sqrt{98} (-2) \sin\theta + (-2)^2 +(5)^2 } = \sqrt{98} $$
Numerically there are two solutions at $\theta=28.27437°$ and $\theta=45°$ for the solutions
$$ (x_0,y_0) = ( -4.6893, 8.7184 ) \\ (x_0,y_0) = ( -7, 7 ) $$
