# Tutorial on Equations of Circles

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:

## 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\( (x - 2)^2 + (y - (- 4))^2 = 5^2 \)

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.

### 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 \( \left( \dfrac{{(-1 + 3)}}{2} , \dfrac{{(2 + 2)}}{2} \right) = C(1,2) \)

The radius r is half the distance between A and B. Hence

\( r = \dfrac{1}{2} \sqrt{ [3 - (-1)]^2 + [2 - 2]^2 } \)

\( = \dfrac{1}{2} \sqrt{4^2 + 0^2} \)

\( = 2 \)

The coordinate of C and the radius r are used in the standard equation of the circle to obtain the equation:

\( (x - 1)^2 + (y - 2)^2 = 2^2 \)

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).

### Example 3

Find the center and radius of the circle with equation \( x^2 - 4x + y^2 - 6y + 9 = 0 \)### 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 \( x^2 \) together and all terms with \( y \) and \( y^2 \) together using parentheses.\( (x^2 - 4x) +( y^2 - 6y) + 9 = 0 \)

We now complete the square within the parentheses.

\( (x^2 - 4x + \color{#FF0000}{4}) - \color{#FF0000}{4} + ( y^2 - 6y + \color{#FF0000}{9}) - \color{#FF0000}{9} + 9 = 0 \)

Which may be written as.

\( (x - 2)^2\ +\ ( y - 3)^2\ - \color{#FF0000}{4} - \color{#FF0000}{9}\ +\ 9 = 0 \)

Simplify and write in standard form

\( (x - 2)^2\ +\ ( y - 3)^2\ =\ 4 \)

\( (x - 2)^2\ +\ ( y - 3)^2\ =\ 2^2 \)

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 3__

Find the center and radius of the circle with equation

\( x^2 - 2x + y^2 - 8 y + 1 = 0 \)

### 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 \( r = \sqrt{9} = 3 \)

distance from C to P is equal to: \( \sqrt{[3 - (-2)]^2 + [4 - 3]^2} \)

\( = \sqrt{5^2 +1^2} \)

\( = \sqrt{26} \)

Since the distance from C to P is \( \sqrt{26} \) 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

\( (x - 1)^2 + ( y + 3)^2 = 4 \)

### 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.

\( d(A,C) = \sqrt{(h - 0)^2 + (k - 4)^2} = r \)

\( d(B,C) = \sqrt{(h - 3)^2 + (k - 5)^2} = r \)

\( d(D,C) = \sqrt{(h - 7)^2 + (k - 3)^2} = r \)

Write that \( d(A,c) = d(B,C) \) and \( d(A,C) = d(D,C) \).

\( \sqrt{(h - 0)^2 + (k - 4)^2} = \sqrt{(h - 3)^2 + (k - 5)^2} \)

\( \sqrt{(h - 0)^2 + (k - 4)^2} = \sqrt{(h - 7)^2 + (k - 3)^2} \)

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.

\( r = \sqrt{(3 - 0)^2 + (0 - 4)^2} \)

\( = 5 \)

The equation of the circle is given by.

\( (x - h)^2 + (y - k)^2 = r^2 \)

\( (x - 3)^2 + y^2 = 25 \)

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

### 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).

\( x + y = 2 \)

\( y = - x + 2 \)

\( m_1 = -1 \)

We now use the formula: \( m_1 \times m_2 = - 1 \) to find the slope m2 of line CT.

\( m_2 = -1 / m_1 = 1 \)

The equation of the line CT which passes by the center C(3 , 5) is given by

\( y - 5 = m_2 (x - 3) \)

\( y = x + 2 \)

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.

\( r = \sqrt{ (3 - 0)^2 + (5 - 2)^2 } = 3\sqrt{2} \)

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:

\( (x - h)^2 + (y - k)^2 = r^2 \)

\( (x - 3)^2 + (y - 5)^2 = (3\sqrt{2})^2 \)

\( (x - 3)^2 + (y - 5)^2 = 18 \)

Shown below is the graph of the circle and the line \( x + y = 2 \) tangent to it.

__Matched Exercise 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.

Tutorials on equation of circle.

Interactive tutorial on equation of circle.

Three Points Circle Calculator.