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?
 A: We can try as follows:
Let $O(h,k)$ be the centre and $r$ be the radius
So, $r=\sqrt{(1-h)^2+(-1-k)^2}$
Again as the distance of a tangent from the centre is equal to the radius
$$r=\dfrac{|3h-4k-10|}5=\dfrac{|4h+3k+5|}5$$
From the last relation, we can find $k$ in terms of $h$
We can replace that value of $k$ in $$\sqrt{(1-h)^2+(-1-k)^2}=\dfrac{|3h-4k-10|}5$$
Take square in both sides to find $h$
Can you take it home from here?
A: As pointed out in comments, the original lines are not parallel to coordinate axes. So if you want them to be coordinate axes, you will need rotation, in addition to shifting of origin. Instead you can find the equation of lines in new coordinate system after shifting the origin and complete the work using your approach.
In my answer, I am sharing another approach for this specific problem -

First, given that both lines are tangent to the circle and intersect at a point, we know that the center of the possible circles is on angle bisector of both lines.
Angle bisector of both lines is given by,
$4x+3y+5 = \pm(3x-4y-10)$ and we get two equations,
$y = - \dfrac{x}{7} - \dfrac{15}{7} \ $ and $ \ y = 7x-5$
Now we know x-intercept of $4x+3y+5=0 \ $ is $ - \dfrac{5}{4}$ and of $3x-4y-10=0$ is $\dfrac{10}{3}$. Given x-coordinate of point $(1, -1)$ being between the two x-intercepts, we know that the slope of angle bisector that we choose has to be between the slopes of tangent lines. That leads to the center of circles being on $y = 7x-5$.
So if radius of circle is $r$ and center is $(h, k)$, equating distance between center and $(1, -1)$,
$(h-1)^2 + (7h-4)^2 = r^2 \ \ $ ...$(i)$
$k = 7h-5 \ \ $ ...$(ii)$
Second we find distance from point $(1, -1)$ to both tangent lines. We have, $|d_1| = \dfrac{6}{5}, |d_2| = \dfrac{3}{5}$
As both tangent lines are perpendicular to each other, we know that (refer diagram for clarification)
$\left(\dfrac{6}{5}- r\right)^2 + \left(\dfrac{3}{5}-r\right)^2 = r^2 \implies r = 3, \dfrac{3}{5}$
Plugging both values of $r$ into $(i)$, we can find values of $h$ and then $k$ from $(ii)$.
Now for each value of $r$, $(i)$ gives us a quadratic in $h$ so that will return two values. We can test which one works. To do that, we can check distance to tangent lines from $(h, k)$ should be equal to radius. But as we know the intersection point of tangent lines is $\left(\dfrac{2}{5}, -\dfrac{11}{5}\right)$, we can simply test
$ \left(h - \dfrac{2}{5}\right)^2 + \left(k + \dfrac{11}{5}\right)^2 = 2 r^2$ should be true (refer diagram for clarification).
Finally we get equation of two circles as,
$(x-1)^2+(y-2)^2 = 9 \ $ and $ \ \left(x-\dfrac{13}{25}\right)^2 + \left(y + \dfrac{34}{25}\right)^2 = \dfrac{9}{25}$
A: An alternative solution to find radius from @Math Lover's solution
if $(h,k)$ is center of circle ,then it's radius is given by the perpendicular distance of $(h,k)$ from any tangent
so,
$$r=\frac{|4h+3k+5|}{5}$$
from equation ($ii$) (see Math Lover's answer)
$$r=\frac{|4h+3(7h-5)+5|}{5}$$
$$ \implies  r=|5h-2| ...(iii)$$
substituting in equation ($i$) we get $h=1,\frac{13}{25}$,from equations ($ii$) and ($iii$) we get  $k,r$ also.I hope you can continue
A: 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}
