Drawing Circles in the Coordinate Plane

Rule

Standard Equation of a Circle

The standard equation of a circle with radius $r$ is the relation between the center of the circle, $(h, k),$ and any point that lies on the circle, $(x, y) .$

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

The distance between the center and any point on the circle is always the radius, $r .$

The equation of a circle can be derived from the Pythagorean Theorem.

Derivation

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

Consider a circle centered at $(h, k)$ such that the point $(x, y)$ lies on the circle.

A right triangle can be created by drawing a horizontal segment to the right of $(h, k)$ and a vertical segment down from $(x, y) .$

The hypotenuse of the triangle has the length $r .$ To find the length of the legs, the coordinates of the third vertex are marked.

The lengths of the legs can now be expressed as the difference between the $x$ - and $y$ -coordinates of the vertices.

Therefore, the length of the legs can be expressed as $(x - a)$ and $(y - b) .$ Since the triangle is a right triangle, the Pythagorean Theorem applies. $a^{2} + b^{2} = c^{2}$ Here, $a$ and $b$ are the lengths of the legs and $c$ is the hypotenuse. Substituting the stated values for the triangle drawn above yields the standard equation of a circle. $(x - h)^{2} + (y - k)^{2} = r^{2}$

Graph the circle from the equation

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Exercise

Graph the circle from the equation $(x + 3)^{2} + (y - 1)^{2} = 2^{2} .$

Show Solution

Solution

If a circle's center and radius are known the circle can be drawn in a coordinate plane by writing the given equation in standard form. $(x - h)^{2} + (y - k)^{2} = r^{2}$ From the given equation, it can be seen that the center of the circle is $(- 3, 1)$ and its radius is $2 .$ First, the center can be marked.

The circle can now be drawn using a compass. Set the compass to the length of $2$ units by placing the needle at the center $C$ and the pencil at any point $2$ units away from $C .$

Now, draw the corresponding circle with the pencil on the compass.

The following graphs shows the circle with the given equation.

Method

Writing the Standard Equation of a Circle

The standard equation for a circle can be written using the radius, the center, and a point on the circle.

The equation of the circle shown above can be written using the following method.

1

State the known values

The general form of the equation of a circle is $(x - h)^{2} + (y - k)^{2} = r^{2},$ where $(h, k)$ is the center of the circle and $r$ is the radius. Of the values needed to write the equation, the center is given: $P (- 2, 1) .$ Thus, $h = - 2$ and $k = 1 .$ The radius of the circle remains to be found.

2

Find the length of the radius

The radius of the circle can be drawn from $P$ to $Q .$

To determine the radius, calculate the distance between the points

P

and

Q .

By substituting the coordinates,

P (- 2, 1)

and

Q (2, - 2),

into the distance formula, the radius can be calculated.

r = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePointsSubstitute

(- 2, 1)

&

(2, - 2)

r = (- 2 - 2)^{2} + (1 - (- 2))^{2}

Evaluate right-hand side

SubNeg

a - (- b) = a + b

r = (- 2 - 2)^{2} + (1 + 2)^{2}

AddSubTermsAdd and subtract terms

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

CalcPowCalculate power

r = 16 + 9

AddTermsAdd terms

r = 25

CalcRootCalculate root

r = 5

Therefore, the radius of the circle is

5

units.

3

Substitute values

The equation can be written by substituting the values of

h, k

and

r

into the rule.

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

SubstituteValuesSubstitute values

(x - (- 2))^{2} + (y - 1)^{2} = 5^{2}

SubNeg

a - (- b) = a + b

(x + 2)^{2} + (y - 1)^{2} = 5^{2}

CalcPowCalculate power

(x + 2)^{2} + (y - 1)^{2} = 25

The equation for the circle is

(x + 2)^{2} + (y - 1)^{2} = 25 .

Prove or disprove that the point lies on the circle

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Exercise

Given a circle with center at $(- 1, 1)$ and the radius $5$ units. Determine if the point $(- 3, 3)$ lies on the circle.

Show Solution

Solution

To determine if the point lies on the circle, we need to find the equation of the circle. Then, the point lies on the circle if and only if it satisfies the equation. The standard equation of a circle is:

(x - h)^{2} + (y - k)^{2} = r^{2},

where

(x, y)

is a point on the circle,

(h, k)

is the center and

r

is the radius. By substituting

(- 1, 1),

the center of the circle, and its radius,

5,

into the rule, the circle's equation can be written.

(x - (- 1))^{2} + (y - 1)^{2} = (5)^{2}

If the point

(- 3, 3)

satisfies the equation, it lies on the circle. Before we test the point, the equation can be simplified.

(x - (- 1))^{2} + (y - 1)^{2} = (5)^{2}

NegNeg

- (- a) = a

(x + 1)^{2} + (y - 1)^{2} = (5)^{2}

PowSqrt

(a)^{2} = a

(x + 1)^{2} + (y - 1)^{2} = 5

Now, we'll substitute the point

(- 3, 3)

to see if it satisfies the equation.

(x + 1)^{2} + (y - 1)^{2} = 5

SubstituteII

x = - 3

,

y = - 3

(- 3 + 1)^{2} + (3 - 1)^{2} = ? 5

Evaluate left-hand side

AddSubTermsAdd and subtract terms

(- 2)^{2} + 2^{2} = ? 5

CalcPowCalculate power

4 + 4 = ? 5

AddTermsAdd terms

8 \neq = 5

Because

8

does not equal

5,

the point

(- 3, 3)

does not satisfy the equation. Thus, the point does not lie on the circle.

Method

Find the Standard Equation of a Circle by Completing Squares

The equation of a circle is not always given in standard form. Consider the following as an example.

x^{2} + y^{2} + 10 x - 4 y = 20

The equation can be written in standard form by completing the square.

1

Identify binomials

The standard equation of a circle consist of two factored perfect square trinomials on the left-hand side. $(x - h)^{2} and (y - k)^{2}$ By completing the square on the $x$ - and $y$ -terms, perfect square trinomials can be created. The terms on the left-hand side of the equation can be rearranged so that the terms with the same variables are next to each other. $x^{2} + y^{2} + 10 x - 4 y = 20 \Leftrightarrow (x^{2} + 10 x) + (y^{2} - 4 y) = 20$ It is necessary to complete the square on $(x^{2} + 10 x)$ and $(y^{2} - 4 y) .$

2

Find the constants to complete the square

The constants needed to complete the square can be found by splitting the second term in the binomials into two factors, where one is

2 .

This is shown for the

x

-terms.

x^{2} + 10 x

SplitIntoFactorsSplit into factors

x^{2} + 2 \cdot 5 x

CommutativePropMultCommutative Property of Multiplication

x^{2} + 2 x \cdot 5

By adding the square of

5

to the binomial, a perfect square trinomial is created.

x^{2} + 10 x + 25

The same procedure is done for the

y

-terms, resulting in

y^{2} - 4 y + 4 .

3

Add the constants to the equation

The constants should now be added to the equation. $(x^{2} + 10 x + 25) + (y^{2} - 4 y + 4) = 20 + 25 + 4$ The terms on the right-hand side can be added. $(x^{2} + 10 x + 25) + (y^{2} - 4 y + 4) = 49$

4

Simplify the equation

The equation can be simplified by factoring the perfect square trinomials on the left-hand side. First the

x

trinomial will be factored, then the

y .

(x^{2} + 10 x + 25) + (y^{2} - 4 y + 4) = 49

FacPosPerfectSquare

a^{2} + 2 a b + b^{2} = (a + b)^{2}

(x + 5)^{2} + (y^{2} - 4 y + 2^{2}) = 49

FacNegPerfectSquare

a^{2} - 2 a b + b^{2} = (a - b)^{2}

(x + 5)^{2} + (y - 2)^{2} = 49

WritePowWrite as a power

(x + 5)^{2} + (y - 2)^{2} = 7^{2}

Therefore, the equation of the circle, written in standard form, is:

(x + 5)^{2} + (y - 2)^{2} = 7^{2} .

The center of the circle can now be identified as

(- 5, 2)

and the radius as

7

units.

Drawing Circles in the Coordinate Plane

Standard Equation of a Circle

Derivation

Example

Graph the circle from the equation

Writing the Standard Equation of a Circle

1

2

3

Example

Prove or disprove that the point lies on the circle

Find the Standard Equation of a Circle by Completing Squares

1

2

3

4