Tutorial on Equation of Circle

Tutorials with detailed solutions to examples and matched exercises on finding equation of a circle, radius and center. Detailed explanations are also provided.

Definition

A circle is the set of points equidistant from a point C(h,k) called the center . The fixed distance r from the center to any point on the circle is called the radius .
The standard equation of a circle with center at C(h,k) and radius r is as follows:
(x - h)2 + (y - k)2 = r2


Examples with Detailed Solutions

Example 1

Find the equation of a circle whose center is at (2, - 4) and radius 5.

Solution to Example 1

Substitute (h , k ) by (2 , - 4) and r by 5 in the standard equation to obtain
(x - 2)2 + (y - (- 4))2 = 52
Simplify
(x - 2)2 + (y + 4)2 = 25
Set h, k and r parameters into this applet and plot the circle. Verify graphically that the equation is that of the circle with the given center and radius.
Matched Exercise 1
Find the equation of a circle whose center is at (2 , - 4) and radius 3.

Solution.

Example 2

Find the equation of a circle that has a diameter with the endpoints given by the points A(-1 , 2) and B(3 , 2).

Solution to Example 2


The center C of the circle is the midpoint of the line segment making the diameter AB. We first use the midpoint formula to find the coordinates of C.
C ( (-1 + 3) / 2 , (2 + 2) / 2) = C(1,2)
The radius r is half the distance between A and B. Hence
r = (1 / 2) √ ( [3 - (-1)]2 + [2 - 2]2 )
= (1/2)√(42 + 02)
= 2
The coordinate of C and the radius r are used in the standard equation of the circle to obtain the equation:
(x - 1)2 + (y - 2)2 = 22
Simplify
(x - 1)2 + (y - 2)2 = 4
Set the h, k and r parameters into this applet and plot the circle. Verify graphically that the equation is that of a circle with the diameter as given above.
Matched Exercise 2
Find the equation of a circle that has a diameter with the endpoints given by A(0 , -2) and B(0 , 2).

Solution

Example 3

Find the center and radius of the circle with equation
x2 - 4x + y2 - 6y + 9 = 0

Solution to Example 3


In order to find the center and the radius of the circle, we first rewrite the given equation into the standard form as given above in the definition. Put all terms with x and x2 together and all terms with y and y2 together using parentheses.
(x2 - 4x) +( y2 - 6y) + 9 = 0
We now complete the square within the parentheses.
(x2 - 4x + 4) - 4 + ( y2 - 6y + 9) - 9 + 9 = 0
Which may be written as.
(x - 2)2  + ( y - 3)2 - 4 - 9 + 9 = 0
Simplify and write in standard form
(x - 2)2 + ( y - 3)2 = 4
(x - 2)2  + ( y - 3)2 = 22
We now compare this equation and the standard equation to obtain.
center at C(h , k) = C(2 , 3)
and radius r = 2 Matched Exercise 1
Find the center and radius of the circle with equation
x2 - 2x + y2 - 8y  + 1 = 0

Solution

Example 4

Is the point P(3 , 4) inside, outside or on the circle with equation
(x + 2)2 + ( y - 3)2 = 9

Solution to Example 4


We first find the distance from the center of the circle to point P.
Using the given equation the center C is at (-2 , 3)
and the radius r = √9 = 3
distance from C to P is equal to: √([3 - (-2)]2 + [4 - 3]2)
= √(52 +12)
= √(26)
Since the distance from C to P is √(26) which approximately equal to 5.1 is greater than the radius r = 3, point P is outside the circle. You can check your answer graphically using this applet
Matched Exercise 4
Is the point P(-1 , -3) inside, outside or on the circle with equation
(x - 1)2  + ( y + 3)2  =  4

Solution

Example 5

Find the equation of the circle such that the three points A(0 , 4), B(3 , 5) and D(7 , 3) are on the circle.

Solution to Example 5

The distance from the center C(h , k) of the circle to each of the points A, B and D is constant and equal to the radius r of the circle. Write three equations stating that these distances are equal to the radius r.
d(A,C) = √[(h - 0)2 + (k - 4)2] = r
d(B,C) = √[(h - 3)2 + (k - 5)2] = r
d(D,C) = √[(h - 7)2 + (k - 3)2] = r
Write that d(A,c) = d(B,C) and d(A,C) = d(D,C).
√[(h - 0)2 + (k - 4)2] = √[(h - 3)2 + (k - 5)2]
√[(h - 0)2 + (k - 4)2] = √[(h - 7)2 + (k - 3)2]
Square each side of each equation.
(h - 0)2 + (k - 4)2 = (h - 3)2 + (k - 5)2
(h - 0)2 + (k - 4)2 = (h - 7)2 + (k - 3)2
Expand the squares in the above equations and simplify.
-8k + 16 = -6h + 9 -10k + 25
-8k + 16 = -14h + 49 -6k +9
Write the above system of equations in standard form.
2k + 6h = 18
-2k + 14h = 42
Use the method of addition to solve the system.
20h = 60
h = 3
Substitute h by its value 6 in one of the equations to obtain k.
k = 0
We now use one of the distance formula in part a above to find the radius r.
r = √[(3 - 0)2 + (0 - 4)2]
= 5
The equation of the circle is given by.
(x - h)2 + (y - k)2 = r2
(x - 3)2 + y2 = 25

Shown below is the graph of the circle with the three points.

graph of circle with the 3 points

Matched Exercise 5
Find the equation of the circle such that the three points A(-5 , 0), B(1 , 0) and D(-2 , -3) are on the circle.

Solution

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).

line tangent to circle.

Figure 1. line tangent to a circle and perpendicular to line through center.
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 = r2
(x - 3)2 + (y - 5)2 = (3 √(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.

Figure 2. Graph of solution: line tangent to a circle.
Matched Exercise 6
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).

More References and links related to the equation of a circle.


Find x and y intercepts of Circles - Calculator: A calculator to calculate the x and y intercepts of the graph of a circle given its center and radius.
Find center and the radius of a Circle: Calculates the coordinates of the center and radius of a circle given its equation.
Match Equations of Circles to Graphs. Excellent interactive activity where equations of circles are matched to graphs.
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