<s>[INST] Equation of the family of circle which touch the pair of lines $x^2-y^2+2y-1=0$ The question was to find the equation of the family of circles which touches the pair of lines, $x^2-y^2+2y-1 = 0$
So I tried as follows:-
The pair of lines is, by factoring the given equation, $x^2-y^2+2y-1 = 0$ is $x+y-1=0$, and $x-y+1=0$.
These are tangent to the required circle, so the center of the circle, (let that be $(h,k)$) must lie upon the angle bisector of these two lines. 
Also, the distance of the center $(h,k)$ from these two lines must equal to the radius (assuming it to be $r$).
How do I use these to get the required equation?
 [/INST] As yuou noted, the centre of the circle must lie on $y=1 \vee x=0$. 
In the first case, we consider the centre to be $C(x_c,1)$ and we have:
$$(x-x_c)^2+(y-1)^2=r^2$$
The radius is the minimal distance from $C$ to the line $y=x+1$ or $y=-x+1$. So we have:
$$r=\frac{|x_c|}{\sqrt{2}}$$
Substituing, we have:
$$(x-x_c)^2+(y-1)^2=\frac{x_c^2}{2}$$
In the second case, we consider the centre to be $C(0,y_c)$ and we have:
$$x^2+(y-y_c)^2=r^2$$
Using the method expained before, we arrive at:
$$x^2+(y-y_c)^2=\frac{(y_c-1)^2}{2}$$
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