# 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

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