This is a continuation of tutorial on equations of circles (1) and
(2)

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 property of the circles that a line through the center C of a circle and the point of tangency T (let us call this line CT) and the line x + y = 2 (let us call this line LT) tangent to the circle are perpendicular (see graph below).

The slopes m1 and m2 of two perpendicular lines are related by the formula: m1 × m2 = - 1. Find the slope m1 of line LT.
x + y = 2
y = - x + 2
m1 = -1
We now use the formula: m1 × m2 = - 1 to find the slope m2 of line CT.
m2 = -1 / m1 = 1
The equation of the line CT which passes by the center C(3 , 5) is given by
y - 5 = m2 (x - 3)
y = x + 2
The point of tangency is the intersection of lines CT and LT and is found by solving the system of equations of the two lines.
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 = √[ (3 - 0)^{2} + (5 - 2)^{2} ] = 3√(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} = (3 √(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).