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 $4x + 3y = 7$.
 A: 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 $r = 5$.  
A: 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
