This is a continuation of tutorial on equations of circles.
Example 6: Find the equation of the circle that is tangent to the line whose equation is given by x + y = 2 and has its center at (3,5).
Solution to Example 6:

The first step is to determine the point of tangency of the circle and the line x + y = 2. Use the fact that line through the center of the circle and the point of tangency (let us call this line LC) and the line x + y = 2 (let us call this line LT) tangent to the circle are perpendicular.

The slopes of two perpendicular lines are related by m1*m2 = 1. Find the slope m1 of line LT and the slope m2 of LC.
x + y = 2
y = x + 2
m1 = 1
m2 = 1/m1 = 1

The equation of the line LC is given by
y  5 = m2(x  3)
y = x + 2

The point of tangency is the intersection of lines LC and LT and is found by solving the system of equations:
x + y = 2
y = x + 2
The point of tangency is at (0 , 2).

The distance between the center of the circle and the point of tangency is equal to the radius r of the circle and is given by.
r = sqrt[ (3  0)^{2} + (5  2)^{2} ] = 3sqrt(2)

Let h and k be the x and y coordinates of the center of the circle and r it radius, the equation of the circle in standard form is given by:
(x  h)^{2} + (y  k)^{2} = r^{2}
(x  3)^{2} + (y  5)^{2} = (3sqrt(2))r^{2}
(x  3)^{2} + (y  5)^{2} = 18
Shown below is the graph of the circle and the line x + y = 2 tangent to it. Matched Exercise: Find the equation of the circle that is tangent to the line whose equation is given by x + 2y = 2 and has its center at (0,5).
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