# 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 $4x + 3y = 7$.

• The center must be equidistant from the point and the lines. – Tony Piccolo Feb 2 '14 at 7:55

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 $r = 5$.

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

• Yes. Thank you. That's my idea alaso. Thaank you! – Rigoo Feb 2 '14 at 9:21
• welcome then,good lucks – dato datuashvili Feb 2 '14 at 9:26
• Wait, how to get the point of intersection of line and circle? – Rigoo Feb 2 '14 at 9:36
• no you dont need it there,just consider that distance from center of circle to lines are equal to each other and also to $(x-2)^2+(y-3)^2=r^2$,will not have $r^2$ and you will simplify things – dato datuashvili Feb 2 '14 at 9:44
• I thought, (2,3) is not a center but it is only a point on a circle. Isn't it? – Rigoo Feb 2 '14 at 10:25