Question: Find equation of circles touching the lines $3x+4y-35=0$ and $4x+3y+14=0$, and passing through the point $(-1,5)$.
Firstly, I was able to infer that there should be two such circles. Secondly, I know that the centre of the circles should pass through the acute angle bisector of the two equations, which is $x+y=3$. However, I am unable to proceed nicely, it is getting algebraically complex for me to solve, I'm looking for a fast solution.
One of my Idea was to take $(a,b)$ as the centre of one of the circles.
And solve the following two equations:
$a+b=3$, as $(a,b)$ lies on the angle bisector.
$(a+1)^2 + (b-5)^2= (\frac{3a+4b-35}{5})^2$, equating the radiuses.
However this is getting too long. I'm wondering if there is a simple solution, Like maybe something involving power of point or some synthetic geometry so that the solving is easier.
It is easy to see that the point at which the lines intersect is $I=(-23,26)$. The circle $c$ centered at $I$ with radius $1$ intersects the line $3x+4y=35$ at two points: $\left(-\frac{119}5,\frac{133}5\right)$ and $\left(-\frac{111}5,\frac{127}5\right)$. Among these two, I will pick the point $I_1=\left(-\frac{111}5,\frac{127}5\right)$, which is the one which is closer to $P=(-1,5)$. And now, let $I_2=\left(-\frac{112}5,\frac{126}5\right)$, which is, among the points where $c$ intersects the line $4x+3y=-14$, the one which is closer to $P$. So, the line defined by $I$ and the midpoint $M$ of the line segment joining $I_1$ and $I_2$ bissects the angle $\angle I_1\hat II_2$. But $M$ is the point $\left(-\frac{223}{10},\frac{253}{10}\right)$ and the slope of the line segment joining this point to $I$ is $-1$. Therefore, the line that bissects the angle $\angle I_1\hat II_2$ is the line $\{(-23+t,26-t)\mid t\in\Bbb R\}$.
So, each point of this line is equidistant to the lines $3x+4y=35$ and $4x+3y=-14$. The distance from $(-23+t,26-t)$ to the first line (and hence to the second line) is$$\frac{|3(-23+t)+4(26-t)-35|}{\sqrt{3^2+4^2}}=\frac{|t|}5.$$
Finally, asserting that $P$ belongs to the circle centered at $(-23+t,26-t)$ with radius $\frac{|t|}5$ means that$$(-23+t+1)^2+(26-t-5)^2=\left(\frac{|t|}5\right)^2,$$whose solutions are $t=25$ and $t=\frac{925}{49}$. So, the circles that you are after are the circle centered at $(2,1)$ with radius $5$ and the circle centered at $\left(-\frac{202}{49},\frac{349}{49}\right)$ with radius $\frac{185}{49}$ (see the picture below).