Find the equation of the circle centered at the point (2, 3) and touching the line 5x+12y-7=0?

1 Answer
Feb 14, 2018

Equation of circle is #(x-2)^2+(y-3)^2=3^2# or #x^2+y^2-4x-6y+4=0#

Explanation:

As #(2,3)# is the center of the circle

and line #5x+12y-7=0# touches it, this is a tangent.

In a circle, length of perpendicular from center to tangent is radius,

hence radius is #|(5xx2+12xx3-7)/sqrt(5^2+12^2)|#

or #|39/13|=3#

Hence, equation of circle is

#(x-2)^2+(y-3)^2=3^2# or #x^2+y^2-4x-6y+4=0#

graph{(x^2+y^2-4x-6y+4)(x^2+y^2-4x-6y+13-0.01)(5x+12y-7)=0 [-9.17, 10.83, -3.88, 6.12]}