Equation of a circle given one point and two lines

Question

Find the equation of the circle that pass through $(2,3)$ and are tangent to both the lines $3x - 4y = -1$ and $4x + 3y = 7$.

Answer 1

i would solve this problem right this

first of equation of this circle is

$(x-2)^2+(y-3)^2=r^2$

now it is tangent of lines

$3*x-4*y+1=0$

$4*x+3*y-7=0$

that means that distance from center to point intersection of circle and lines must be equal to each other,in other word ,distance from center to $1$ line and distance from center to another line must be equal to each other,

distance formula from point $(x_0,y_0)$ to line $A*x+b*y+c=0$

is

$(A*x_0+b*y_0+c)/\sqrt{A^2+b^2}$

could you continue from this?also please pay attention that this distance is the same as radius

Answer 2

Using the angle bisector equation, we obtain that the center of the required circle has to lie on the line $\frac{3x-4y+1}{5} = \pm \frac{4x+3y-7}{5}$. Draw a figure and you can see that we need to consider the equation with minus on RHS. Hence, the required angle bisector $L : 7x-y-6=0$.Or, the center is of the form $(x, 7x-6)$.

We know that the center has to be equidistant from the point $(2,3)$ and both the lines. This condition gives $(x-2)^2 + (7x-6-3)^2 = \frac{(3x - 4(7x- 6) + 1)^2}{25}$ which on simplification gives $x = 2 \text{ or } 6/5$. You can easily verify from the diagram that the required point is $(2,8)$. And the radius $r = 5$.