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?
Let $(u, v)$ is a center of a circle.

Since both lines $x+y-1 = 0$ and $x-y+1 = 0$ are the tangent line, the distance from $(u, v)$ to these lines must be the same.

By using the formula of the distance from a point to a line, we have the following equation:

$$
r = \frac{|u-v+1|}{\sqrt{2}} = \frac{|u+v-1|}{\sqrt{2}} \Leftrightarrow |u-v+1| = |u+v-1|.
$$

Here, $r$ is the radius of a correponding circle.

There are four possible scenarios:

Scenario 1. 

$$
\left\{
\begin{aligned}
u-v+1 &amp;&gt; 0 \\
u+v-1 &amp;&gt; 0 \\
u-v+1 &amp;= u+v-1
\end{aligned}
\right. \Leftrightarrow (u &gt; 0, v = 1)
$$

The radius of such a circle equals to 
$$
r = \frac{|u|}{\sqrt{2}},
$$

and the corresponding family of circles is:

$$
(x-u)^2 + (y-1)^2 = \frac{u^2}{2}, \text{ }u &gt; 0.
$$

Scenario 2. 

$$
\left\{
\begin{aligned}
u-v+1 &amp;&lt; 0 \\
u+v-1 &amp;&gt; 0 \\
-(u-v+1) &amp;= u+v-1
\end{aligned}
\right. \Leftrightarrow (u = 0, v &gt; 1)
$$

The radius of such a circle equals to 
$$
r = \frac{|v-1|}{\sqrt{2}},
$$

and the corresponding family of circles is:

$$
x^2 + (y-v)^2 = \frac{(v-1)^2}{2}, \text{ }v &gt; 1.
$$

Scenarios 3 and 4  I left to you.

Scenario 3. 

$$
\left\{
\begin{aligned}
u-v+1 &amp;&gt; 0 \\
u+v-1 &amp;&lt; 0 \\
u-v+1 &amp;= -(u+v-1)
\end{aligned}
\right. 
$$

Scenario 4. 

$$
\left\{
\begin{aligned}
u-v+1 &amp;&lt; 0 \\
u+v-1 &amp;&lt; 0 \\
-(u-v+1) &amp;= -(u+v-1)
\end{aligned}
\right. 
$$
EDIT1:

Factorizing gives two straight lines

$$y-1=-x,\, y-1=+x $$

Depressing the pair of straight lines by $1.0$ for all y-coordinate results the pair  at $\pm 45^{\circ}$ slope easy to handle:

$$  (x+y)(x-y)=0 $$

whose touching lines can be readily rough sketched as shown below with variable (x, y) axes displacements $(h,k)$ for location of circle centers.



Now elevate the lines back by same amount $1$

$$y\rightarrow y+1 $$

to obtain required new  $(h,k)$ parameter set of circle equations touching the line pair:

$$ (x-h)^2 + (y-1)^2 = h^2/2; \quad  x^2+(y-1-k)^2= k^2/2.$$