Find the distance from (3,5) to the line specified. That is your radius.

the orthogonal slope of the intersection will be given by the negative inverse of tangent slope, or -5/12x. Now that you know that, just calculate the intersection of (y-5)= -5/12(x-3) and 12x - 5y +2=0. Once you have the intersection, you can then calculate the radius, and then you know that (x-3)^2 + (y-5)^2 = r^2.

The standard circle equation is: (x-a)^2 + (y-b)^2 = r^2 where (a,b) is the center point and r is the radius. You know the center point (3,5) but you must find the radius. To find the radius, you must find the point that lies on the circle at the tangent line. To do this, you must find the line that passes through the center of the circle and then find the intersection of that line and the tangent line. The distance between the two points will be the radius. To find the line that passes through the center of the circle do the following Rewrite the tangent line into the form: y = (m1)x + b1 Let the perpendiculer line be: y = (m2)x + b2 You find m2 by the equation: m2 = -1/m1 Now that you know m2, find b2 by substituting the center point of the circle into the equation: 5 = m2(3) + b2 b2 = 5 - 3(m2) To find the point (xp,yp) you find the intersection of the two lines: y = (m1)x + b1 y = (m2)x + b2 To find the x value of the intersection point you set the right hand sides equal to each other. (m1)x + b1 = (m2)x + b2 (m1)x = (m2)x + b2 - b1 {m1 - m2}x = b2 - b1 The x value of the intersection point is: xp = (b2 - b1)/(m1 - m2) To find yp substitute xp back into the equation: 12x - 5y + 2 = 0 and solve for y. Now that you have a point on the circle, you can substitute back into the equation of the circle (which is the same as computing the distance between two points but stopping before you take the squareroot.). (xp-3)^2 + (yp-5)^2 = r^2 This will give you the value of r^2.