<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] 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 $\frac{3}{2}$ $$\left(x-\frac{3}{2\sqrt{2}}\right)^2+\left(y-\frac{1}{2\sqrt{2}}\right)^2=\left(\frac{3}{2}\right)^2 $$$$\implies \color{#0b4}{4x^2+4y^2-6x\sqrt{2}-2y\sqrt{2}-4=0}$$
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