Tutorial on Equation of Circle(3)

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).

• Tutorials on equation of circle.
  
  
• Tutorials on equation of circle (2).
  
  
• Interactive tutorial on equation of circle.