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
 A: 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
$$
\left(x - \frac{3}{2\sqrt{2}} \right)^2 + 
\left(y - \frac{1}{2\sqrt{2}} \right)^2 = 9/4
$$

A: hints:
Since $\;(0,\,\sqrt2)\;$ is both on the circle and on one of the lines (why and which line?), the line must be perpendicular to the radius of the circle at this point (why?), so the circle's center is on the line 
$$y-\sqrt2=-(x-0)\implies y=-x+\sqrt2$$
Since the given second line is parallel to the given first one (why?), these two are tangent to point of the circle on end points of a diameter (why?).
End the exercise now.
A: Let the given point be $A(0, \sqrt{2})$ which lines on the lies on the line: $y=x+\sqrt{2}$, Now draw the perpendicular say $AN$ from the point $A(0, \sqrt{2})$ to the parallel line: $y=x-2\sqrt{2}$ at the point N. 
Thus point N is foot of perpendicular & $AN$ denotes the diameter of circle. Point $N$ is determined by a general expression derived in Reflection Formula by HCR for directly calculating the co-ordinates of foot of perpendicular say $(x', y')$ drawn from any point $(x_o, y_o)$ to the straight line $y=mx+c$ given as $$(x', y')\equiv\left(\frac{x_o+m(y_o-c)}{1+m^2}, \frac{mx_o+m^2y_o+c}{1+m^2}\right)$$ Hence, the co-ordinates of the foot of perpendicular $N$ drawn from the point $A(0, \sqrt{2})$ to the line: $y=x-2\sqrt{2}$ are calculated as follows $$N\equiv\left(\frac{0+1(\sqrt{2}-(-2\sqrt{2}))}{1+1^2}, \frac{1(0)+1^2(\sqrt{2})-2\sqrt{2}}{1+1^2}\right)\equiv\left(\frac{3}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right)$$ Now, the center of circle is the mid-point of the line AN joining $A(0, \sqrt{2})$ & $N\left(\frac{3}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right)$ which is given as $\left(\frac{3}{2\sqrt{2}}, \frac{1}{2\sqrt{2}}\right)$ & radius of circle is $\frac{3}{2}$ i.e. half the distance ($=3$) between parallel lines.
Hence, the equation of the circle with center $\left(\frac{3}{2\sqrt{2}}, \frac{1}{2\sqrt{2}}\right)$ & radius $\frac{3}{2}$ $$\left(x-\frac{3}{2\sqrt{2}}\right)^2+\left(y-\frac{1}{2\sqrt{2}}\right)^2=\left(\frac{3}{2}\right)^2 $$$$\implies \color{#0b4}{4x^2+4y^2-6x\sqrt{2}-2y\sqrt{2}-4=0}$$
A: Assume the center to be (a,b). Write the general equation of the circle $(x-a)^2+(y-b)^2=(3/2)^2$ and substitute the point $(0, \sqrt{2})$. This is the first equation.
It is also true that the point (a,b) lies on the the line given by $y = x - \dfrac{\sqrt{2}}{2}$. Substitute for 'a' and 'b'. This is the second equation.
Now you have two unknowns and two equations. Solve it!
