<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] 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}$
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