<s>[INST] Coordinates of the center of the circle I am stuck on this problem:

If the lines $y=x+\sqrt{2}$ and $y=x-2\sqrt{2}$ are two tangents of a
   circle and $(0,\sqrt{2})$ lies on this circle then what is the equation of the circle?

I found out the distance between the two  tangents  $y=x+\sqrt{2}$ and $y=x-2\sqrt{2}$ is $3$. The radius is $3/2$ but I don't know how to find the center. I tried forming the equations but could not succeed. Please tell me the easiest way. Thank you
 [/INST] Hint: A sketch of the situation should be helpful:

What can we say about the position of the center of a circle with those two tangents?
What is its radius?
The equation of a circle with radious $r$ and center $(x_0,y_0)$ is:
$$
(x - x_0)^2 + (y - y_0)^2 = r^2
$$
Solution:
a) Center line: The center points lie on a line in the middle of the lines, thus on
\begin{align}
y_c(x) 
&= \frac{1}{2}(y_1(x) + y_2(x)) \\
&= \frac{1}{2}((x + \sqrt{2}) + (x - 2 \sqrt{2})) \\
&= x - \frac{1}{\sqrt{2}}
\end{align}
b) Radius: The radius of the circle is two times the distance of the tangent lines.
An orthogonal line to both tangents is $y = - x$ it intersects $y_1$ if 
$$
x + \sqrt{2} = - x \iff 2x = -\sqrt{2} \iff x = - 1/\sqrt{2}
$$
thus at $P = (-1/\sqrt{2}, 1/\sqrt{2})$. 
The distance to the origin is $\lVert OP \rVert = \sqrt{1/2 + 1/2} = 1$
It intersects $y_2$ if
$$
x - 2 \sqrt{2} = -x \iff 2x = 2 \sqrt{2} \iff x = \sqrt{2}
$$
thus at $Q = (\sqrt{2}, -\sqrt{2})$. 
The distance to the origin is $\lVert OQ \rVert = \sqrt{2 + 2} = 2$.
So both tangents are at a distance $3$ and the radius is $3/2$.
c) Equation of any circle with those tangents:
So all circles with those two tangents have the equation
$$
(x - x_0)^2 + \left(y - x_0 + \frac{1}{\sqrt{2}}\right)^2 = 9/4
$$
d) The circle with those tangents that contains $(0, \sqrt{2})$:
We want the one that contains $(0, \sqrt{2})$ thus
$$
9/4 = x_0^2 + (\sqrt{2} - x_0 + 1/\sqrt{2})^2
= x_0^2 + (3/\sqrt{2} - x_0)^2
= x_0^2 + 9/2 + x_0^2 - 3\sqrt{2} x_0 \Rightarrow \\
-9/4 = 2 x_0^2 - 3\sqrt{2} x_0 \Rightarrow \\
-9/8 = x_0^2 - (3/\sqrt{2}) x_0 
= (x_0-3/(2\sqrt{2}))^2 - 9/8 \Rightarrow \\
x_0 = \frac{3}{2\sqrt{2}}
$$
This gives the center 
$C=(3/(2\sqrt{2}), 3/(2\sqrt{2}) - 1/\sqrt{2}) = (3/(2\sqrt{2}), 1/(2\sqrt{2}))$.
And we have the equation
$$
\left(x - \frac{3}{2\sqrt{2}} \right)^2 + 
\left(y - \frac{1}{2\sqrt{2}} \right)^2 = 9/4
$$

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