The equation of the line cutting the circle at $A$ and $B$ given that $AB=\sqrt{2}$ Given the circle $C: 2x^2 + 2y^2 + 2x + 2y - 13 = 0$. Line $L$, with slope $m$ and passing through the point $P(0,2)$, cuts the circle at points $A$ and $B$ such that $AB=\sqrt{2}$. 


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*Find the equation of $L$.

*Find the equation of the locus of the centers of the circles passing through $A$ and $B$.


The equation of the line is obviously $L:y=mx+2$
The length of the chord $AB$ is $\sqrt{2}$ so the following must be satisfied:
$(x_B-x_A)^2+(y_B-y_A)^2 = 2$
I tried a few things to calculate $m$, but unsuccessfully.
I guess that once I find $m$ the equation of the locus of the centers of the circles passing through $A$ and $B$ is calculated based on the condition that the locus is perpendicular to $AB$ (please correct me if I am wrong).
Any hint would be useful.

Let $x$ the distance from the centre of the circle $(x+1/2)^2+(y+1/2)^2=7$, I have $x=\sqrt{7-\frac{1}{2}}=\frac{\sqrt{13}}{\sqrt{2}}$. The distance of the line from the centre $P(-1/2,-1/2)$ has to be equal to $x$, so I obtain: $$\frac{|\frac{m}{2}-\frac{1}{2}-2|}{\sqrt{m^2-4}}=\frac{\sqrt{13}}{\sqrt{2}}$$ Solving for $m$ I have that: $25m^2+10m+1=0$ that has only one solution when $m=-\frac{1}{5}$. The equation of the line is: $$y=-\frac{1}{5}x+2$$
