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?
 A: 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. 
$$
\left\{
\begin{aligned}
u-v+1 &> 0 \\
u+v-1 &> 0 \\
u-v+1 &= u+v-1
\end{aligned}
\right. \Leftrightarrow (u > 0, v = 1)
$$
The radius of such a circle equals to 
$$
r = \frac{|u|}{\sqrt{2}},
$$
and the corresponding family of circles is:
$$
(x-u)^2 + (y-1)^2 = \frac{u^2}{2}, \text{ }u > 0.
$$
Scenario 2. 
$$
\left\{
\begin{aligned}
u-v+1 &< 0 \\
u+v-1 &> 0 \\
-(u-v+1) &= u+v-1
\end{aligned}
\right. \Leftrightarrow (u = 0, v > 1)
$$
The radius of such a circle equals to 
$$
r = \frac{|v-1|}{\sqrt{2}},
$$
and the corresponding family of circles is:
$$
x^2 + (y-v)^2 = \frac{(v-1)^2}{2}, \text{ }v > 1.
$$
Scenarios 3 and 4  I left to you.
Scenario 3. 
$$
\left\{
\begin{aligned}
u-v+1 &> 0 \\
u+v-1 &< 0 \\
u-v+1 &= -(u+v-1)
\end{aligned}
\right. 
$$
Scenario 4. 
$$
\left\{
\begin{aligned}
u-v+1 &< 0 \\
u+v-1 &< 0 \\
-(u-v+1) &= -(u+v-1)
\end{aligned}
\right. 
$$
A: I propose this approach (I will find circles in the right sector, for the others sectors the computation is really similar):
Choose a point $A_a=(a,a+1)$ ($a>0$) that lies in the line $r=\{(x,y) \ | \ y=x+1\}$. 
Now draw the line passing through the point $A_a$ and it is orthogonal to the line $r$: with a simple computation you find $s_a=\{(x,y) \ | \ y = -x + 2a+1\}$. In this line will lie the radius of the circle we are looking for.
Take the intersection point between $s_a$ and the right bisector (that has equation $y=1$). You find the point $C_a = (2a,1)$. This will be the center of our circle.
Find the distance $\overline{A_aC_a}$ that is $r_a = \sqrt{(a-2a)^2 + (a+1-1)^2}=a\sqrt 2$; this is the length of radius.
Now we have all the element to write the equation of the circle; the points lie in the circle are points $P$ such that $\overline{PC_a} = r_a^2$ so the equation is:
\begin{gather}
(x-2a)^2+(y-1)^2 = (a\sqrt 2)^2\\
x^2+y^2-4ax-2y+2a^2+1 =0
\end{gather}
A: 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:
$$x^2+(y-y_c)^2=\frac{(y_c-1)^2}{2}$$
A: 1) $y=-x+1$; 2)$y=x+1$;
Both lines have $y-$intercept $1$, and intersects at $(0,1).$
2) Angle bisectors :
$y=1$; and $x=0$;
Note: The lines make angles of $\pm 45°$ with the horizontal.
3) For the first quardrant and $y=1$: 
The centres of the circles are at $x=t \ge 0$, $y=1$ i.e $C(t,1)$.
$r=t \sin 45°=(1/2)√2t$.
$(x -t)^2-(y-1)^2=(1/2)t^2$;
4) Consider centres of the circles $C(0,t+1)$, $t\ge 0$, i.e. along $y-$axis, $y \ge 1$.
$x^2+(y-(t+1))^2 =(1/2)t^2$.
Can you finish?
A: The point-line distance formula for the point $(h,k)$ and the line $ax+by+c=0$ is
$$d= \frac{|ah+bk+c|}{\sqrt{a^2+b^2}}$$
Apply the distance formula to the circle centers $(0,p)$, $(p,1)$ on the respective angle bisectors and the tangent line $x\pm y\mp1=0$ to calculate the two sets of radii,
$$r= \frac{|\pm p \mp1|}{\sqrt{2}}=  \frac{|p-1|}{\sqrt{2}},\>\>\>\>\>
r= \frac{|p \pm 1\mp1|}{\sqrt{2}}=  \frac{|p|}{\sqrt{2}}
$$
Thus, the two families of the circles are
$$x^2+(y-p)^2= \frac12(p-1)^2,\>\>\>\>\>(x-p)^2+(y-1)^2= \frac12p^2$$
A: 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:
$$ (x-h)^2 + (y-1)^2 = h^2/2; \quad  x^2+(y-1-k)^2= k^2/2.$$
