Tutorial on Equation of Circle(3)
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 = r2
(x - 3)2 + (y - 5)2 = (3sqrt(2))r2
(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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Updated: 2 April 2013
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