# Finding The Equation of A Circle

Finding The Equation of A Circle

A circle is a set of all points in a plane equidistant from a fixed point which is the center of the circle. The distance from the center to a point on the circle is called the radius.

Standard equation for a circle with center at the origin : X^2 + Y^2 = r^2

Standard equation for a circle with center at (h, k) : (X – h)^2 + ( Y – k)^2 = r^2

General equation of the circle : X^2 + Y^2 + DX + EY + F = 0.

**Sample Problems
Involving Circle :**

**Problem Number One :**

**Find the equation
of a circle which passes through points
(10, 2) ; (3, 9 ); (-2. 10).**

Solution : Consider the general equation of a circle as X^2 + Y^2 + DX + EY + F = 0.

Using the given three points we derive our equation :

From the point ( 10, 2 ) we get the equation 100 + 4 + 10D + 2E + F = 0 or

104 + 10D + 2E + F = 0 by substituting 10 to X’s and 2 to Y’s . This is our equation one.

From the point (3, 9) we get 9 + 81 + 3D + 9E + F = 0 or 90 +3D + 9E + F = 0.

From the point (-2.10) we get 4 + 100 – 2D + 10E + F = 0 or 104 -2D+ 10E +F = 0.

We now have a system of three equations :

104 + 10D + 2E + F = 0 eqn one

90 + 3D + 9E +F = 0 eqn two

104 = 2D + 10E + F = 0 eqn three

We now use elimination method to find the value of D, E and F.

Using equation one and eqn two to eliminate F, we do this by subtracting equation two from eqn one :

104 + 10D + 2E + F = 0

-(90 +2D + 9E + F = 0)

We get 14 + 7D - 7E = 0 let this be eqn four.

Using Equation 2 and eqn 3 to eliminate F we subtract eqn 3 from eqn 2 :

90 + 3D + 9E + F = 0

- ( 104D -2D +10E + F =0)

We get -14 + 5D - E = 0 let this be equation five

Using eqn four and eqn five we can solve for D and E.

14 + 7D - 7E = 0 eqn four

-14 +5D - E = 0 eqn five

Multiple equation five by -7 in order to eliminate E

(-14 + 5D -
E = 0) *
(-7) =è 98 _{}=35D + 7E = 0

Adding eqn 4 and eqn 5

14 + 7D - 7E = 0

98 - 35D + 7E = 0

112 -28D = 0

(1/28) 112 = 28D (1/28)

D = 4

Substitute D = 4 in equation four ;

14 + 7(4) - 7E = 0

14 + 28 - 7E = 0

42 = 7E

E = 6

Substitute D = 4 , E = 6 in eqn 1 to solve for F

104 + 10(4) + 2(6) + F = 0

104 + 40 + 12 + F = 0

156 + F = 0

F = - 156

The general equation of the circle we are solving is :

X^2 + Y^2 +4X + 6Y-156 = 0.

We get this general equation by substituting D = 4, E = 6 and F = -156 to the general equation of the circle X^2 + Y^2 + DX + EY + F = 0.

To solve for the standard equation of the circle we will use completing the square:

X^2 + 4X + ____ + Y^2 + 6Y + ___ = 156

To complete the square divide the coefficient of the middle term by two then square it.

X^2 + 4X + 4 + Y^2 + 6Y + 9 = 156 + 4 + 9.

The standard equation of the circle we are looking for is :

(X + 2)^2 + (Y + 3)^2 = 169

The circle has its center at (-2, -3) and has a radius equal to 13.

**Problem Number Two :**

**Find the equation of
a circle whose diameter has its
endpoints at A(-3, 5) and B(1, 3).**

To find the center of the circle find the midpoint of the diameter using midpoint formula.

Here is the midpoint formula :

X = (X1 + X2 )/2 is used in finding the X-coordinate of the midpoint.

Y = (Y1 + Y2 )/2 is used in finding the Y-coordinate of the midpoint.

Using the endpoints (-3, 5) and (1, 3)

X = (-3 + 1)/2 = -1 Y = (5 +3 )/2 = 4

Therefore the midpoint of the diameter which is also the center of the circle we are looking for is ( -1, 4)

To find the radius of the circle find the distance between the center and one endpoint using distance formula. The distance formula is :

D = SQRT((Y2 – Y1)^2 + (X2 – X1)^2).

Using one endpoint (1, 3) and the center of the circle ( -1, 4), let us find the distance using the distance formula :

D = SQRT( ( 3-4)^2 + (1 + 1)^2 ) = > SQRT(5)

Therefore the standard equation of the circle we are looking for is :

(X + 1 )^2 + ( Y – 4 )^2 = 5

.

## More by this Author

- 3
Solving Number Problems Involving System of Equations In Two Variables One of the most common applications of equations are number problems. This hub presents five number problems involving system of equations in...

- 4
Solving Geometry Problems Involving System of Equations In Two Variables Algebra has also found wide applications in Geometry. Among its common problems are problems involving perimeters and angles. In this hub I...

- 4
Solving Word Problems Involving Chebyshev’s Theorem Chebyshev’s Theorem states that the proportion or percentage of any data set that lies within k standard deviation of the mean where k is any...

## Comments 9 comments

ah ok

shouldn't you work for NASA or something?

Ok Christina: I can add, subtract, multiply and divide then I am lost. Happy New Year.

there a entries(coefficients, signs)that sometimes confuses me.,but your hub helped me a lot in my analytic geometry assignment., thanks a lot!Godbless!

this helped me a lot my analytic geometry

TY GBY

thanx......it helps a lot for my assignment in advance algebra....and analytic geometry,can i ask more?jejejeje,if its ok.........

this is very helpful thanks

can you write about solving circle using equation?

thanks :))

9