<s>[INST] 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.
 [/INST] Circle: $(x+0.5)^2 + (y+0.5)^2 = 7 = r^2$ 
Distance P (0,2) to center C (-0.5,-0.5) = $\sqrt{0.5^2 + 2.5^2} = \sqrt {6.5}$ 
Distance AB to center = $\sqrt{r^2 - (\frac{AB}{2})^2} = \sqrt{7-{2 \over 4}}= \sqrt {6.5}$ 
Thus, AB perpendicular to line PC, with slope, $m = -\frac{0.5}{2.5} = -0.2$ 
Line L: $y = -x/5 + 2$ 
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