<s>[INST] Find the equations of circles passing through $(1, -1)$ touching the lines $4x+3y+5=0$ and $3x-4y-10=0$ 
Find the equations of circles passing through $(1,-1)$ touching the lines $4x+3y+5=0$ and $3x-4y-10=0$

The point of intersection of the lines is $(\frac25,-\frac{11}5)$
If we want this point of intersection to be $(0,0)$ (because the given lines are perpendicular) then define the new $X=x-\frac25,Y=y+\frac{11}5$, where $x,y$ are the original coordinates.
So, the point $(1,-1)$ in the new system becomes $(\frac35,\frac65)$. Since it's in the first quadrant, so, the center of the circle will be $(r,r)$ where $r$ is the radius of the circle.
So, $(\frac35-r)^2+(\frac65-r)^2=r^2\implies r=\frac35,3$
So, the equations of circles are $$(X-\frac35)^2+(Y-\frac35)^2=\frac9{25}\implies(x-1)^2+(y+\frac85)^2=\frac9{25}$$
and $$(X-3)^2+(Y-3)^2=9\implies(x-\frac{17}5)^2+(y-\frac45)^2=9$$
Is this correct?
 [/INST] Geometric approach.
The centers of all inscribed circles are located on a bisector line,
which equation is $y=7x-5$.
Construct auxiliary inscribed circle, say centered at $O=(2,9)$
with the radius $r=8$.

The line $XP$ intersect this circle at
points $D(2,1)$ and $E(\tfrac{42}5,\tfrac{49}5)$
and we have two scaling coefficients,
\begin{align}
k_1&=\frac{|XP|}{|XE|}
=\tfrac3{40}
,\\
\text{and }\quad
k_2&=\frac{|XP|}{|XD|}
=\tfrac38
,
\end{align}
so we can find the centers and the radii of sought circles through $P$
as
\begin{align}
O_1&=X+(O-X)\cdot k_1
=(\tfrac{13}{25},-\tfrac{34}{25})
,\\
r_1&=r k_1=\tfrac35
,\\
O_2&=X+(O-X)\cdot k_2
=(1,2)
,\\
r_2&=r k_2=3
.
\end{align}
Then the circle equations would be
\begin{align}
(x-\tfrac{13}{25})^2+(y+\tfrac{34}{25})^2&=\tfrac9{25}
\\
\text{and }\quad
(x-1)^2+(y-2)^2&=9
.
\end{align}
</s>