The equation of the line cutting the circle at $A$ and $B$ given that $AB=\sqrt{2}$

Question

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 $AB=\sqrt{2}$.

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

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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.

Answer 1

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: $y = -x/5 + 2$

Answer 2

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: $$y=-\frac{1}{5}x+2$$