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

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
