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.

graph of circle and line 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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