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

  • $\begingroup$ The center must be equidistant from the point and the lines. $\endgroup$ – 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


now it is tangent of lines



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 $A*x+b*y+c=0$



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

  • 1
    $\begingroup$ Yes. Thank you. That's my idea alaso. Thaank you! $\endgroup$ – Rigoo Feb 2 '14 at 9:21
  • $\begingroup$ welcome then,good lucks $\endgroup$ – dato datuashvili Feb 2 '14 at 9:26
  • $\begingroup$ Wait, how to get the point of intersection of line and circle? $\endgroup$ – Rigoo Feb 2 '14 at 9:36
  • $\begingroup$ 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 $\endgroup$ – dato datuashvili Feb 2 '14 at 9:44
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    $\begingroup$ I thought, (2,3) is not a center but it is only a point on a circle. Isn't it? $\endgroup$ – Rigoo Feb 2 '14 at 10:25

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