Equation of a Circle

Tearrik has one last problem to solve before going to a BBQ. He needs to find the standard equation of the circle shown below.

Tearrik remembers that the standard equation of a circle is (x-h)^2+(y-k)^2=r^2. However, he does not remember how to find the values of h, k, and r. Help Tearrik get to the BBQ by finding these values!

Solution

In the standard equation of a circle, (h,k) is the center of the circle. Therefore, by noting the coordinates of the center in the diagram, the values of h and k can be determined. Furthermore, the value of r is given by the distance between the center and any point on the circle.

The center of the circle can be seen to have the coordinates ( - 1, 2). Therefore, it can be stated that h= - 1 and k= 2. It can also be seen that the radius of the circle is r= 5. By substituting these values into the standard equation of a circle, the equation of the given circle can be obtained.

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

SubstituteValues

Substitute values

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

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Simplify

SubNeg

a-(- b)=a+b

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

CalcPow

Calculate power

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

Just like her classmate, Zain has one last problem to solve before getting to go to the BBQ. She has been asked to identify the center and the radius of the circle whose standard equation is given below. (x-4)^2+(y+3)^2=25 Zain has also been asked to graph the circle on a coordinate plane. Help Zain get to the BBQ!

Solution

The standard equation of a circle is shown below. (x-h)^2+(y-k)^2=r^2 Here, the center of the circle is (h,k) and its radius is r. Therefore, it is convenient to rewrite the given equation to match this format.

(x-4)^2+(y+3)^2=25

▼

Rewrite 3 as -(- 3) and Rewrite 25 as 5^2

a+b=a-(- b)

(x-4)^2+(y-(- 3))^2=25

WritePow

Write as a power

(x- 4)^2+(y-( - 3))^2= 5^2

From the obtained equation, the center of the circle can be identified as ( 4, - 3) and its radius as 5. With this information, the circle can be drawn on a coordinate plane.

Sometimes the equation of a circle needs a significant change to be rewritten as the standard equation of a circle. Typically in those cases, the equation can be rewritten by completing the square.

To be allowed to help design her schools basketball court, Dominika was asked to identify the center and the radius of the circle whose equation is given below. x^2-2x+y^2=4 By rewriting the above equation as the standard equation of a circle, identify the center and the radius. If any answer is an irrational number, write its exact value.

Solution

To rewrite the given equation as the standard equation of a circle, a method called completing the square can be used. This method consists in writing an expression as a perfect square trinomial so that it can be factored as the square of a binomial. First, the expression x^2-2x will be considered. x^2-2x+y^2=4 Ignoring the negative sign, the coefficient of the variable x is 2. This coefficient is commonly called b. x^2- 2x+y^2=4 Next, the term ( b2)^2 will be added to and subtracted from the expression. Note that adding and subtracting the same value does not modify the expression. Since b= 2, the value of ( 22)^2 can be calculated.

(2/2)^2

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Simplify

QuotOne

a/a=1

1^2

BaseOne

1^a=1

1

Therefore, 1 will be added to and subtracted from x^2-2x. Then, the resulting perfect square trinomial will be factored and written as the square of a binomial.

x^2-2x+y^2=4

IdPropAdd

Identity Property of Addition

x^2-2x+0+y^2=4

Rewrite

Rewrite 0 as 1-1

x^2-2x+ 1- 1+y^2=4

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a^2-2ab+b^2=(a-b)^2

IdPropMult

Identity Property of Multiplication

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

WritePow

Write as a power

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

FacNegPerfectSquare

a^2-2ab+b^2=(a-b)^2

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

The process of completing the square is now finished. Finally, to obtain the standard equation of the circle, the number 1 will be added to both sides of the equation and y will be written as y-0. Also, the resulting number on the right-hand side will be expressed as a square.

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

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Simplify

AddEqn

LHS+1=RHS+1

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

Rewrite

Rewrite y as y-0

(x-1)^2+(y-0)^2=5

Rewrite

Rewrite 5 as (sqrt(5))^2

(x- 1)^2+(y- 0)^2=( sqrt(5))^2

The standard equation of the circle was obtained. The center can be identified as ( 1, 0) and the radius as sqrt(5).

This time, Dominika wants to play basketball on Sunday. Her father will be okay with that only if she completes her math homework. To do so, Dominika has to identify the center and the radius of the circle whose equation is given below. x^2+6x+y^2-4y=- 3 By rewriting the above equation as the standard equation of a circle, identify the center and the radius. If any of the answers is an irrational number, write its exact value.

Solution

To rewrite the given equation as the standard equation of a circle, the process of completing the square needs to be performed for both variables.

x^2+6x+y^2-4y=- 3

IdPropAdd

Identity Property of Addition

x^2+6x+0+y^2-4y+0=- 3

Rewrite

Rewrite 0 as 9-9 & 4-4

x^2+6x+ 9- 9+y^2-4y+ 4- 4=- 3

▼

a^2± 2ab+b^2=(a± b)^2

SplitIntoFactors

Split into factors

x^2+2(3)x+9-9+y^2-2(2)y+4-4=- 3

CommutativePropMult

Commutative Property of Multiplication

x^2+2x(3)+9-9+y^2-2y(2)+4-4=- 3

WritePow

Write as a power

x^2+2x(3)+3^2-9+y^2-2y(2)+2^2-4=- 3

a^2± 2ab+b^2=(a± b)^2

(x+3)^2-9+(y-2)^2-4=- 3

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Simplify

SubTerms

Subtract terms

(x+3)^2+(y-2)^2-13=- 3

AddEqn

LHS+13=RHS+13

(x+3)^2+(y-2)^2=10

a+b=a-(- b)

(x-(- 3))^2+(y-2)^2=10

Rewrite

Rewrite 10 as (sqrt(10))^2

(x-( - 3))^2+(y- 2)^2=( sqrt(10))^2

The standard equation of the circle was obtained. The center can be identified as ( - 3, 2) and the radius as sqrt(10).

The challenge presented at the beginning of this lesson can be solved by writing the equation of the circle.

On a coordinate plane, a circle centered at the origin with radius 5 was drawn. Also, a point on the circle with x-coordinate 1 was plotted.

By writing the standard equation of the circle, find the y-coordinate of P. Write the answer as an exact value.

Solution

Recall the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 Here, ( h, k) is the center and r the radius of the circle. Knowing that the center of the circle drawn in the diagram is ( 0, 0) and that its radius is 5, its standard equation can be written.

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

SubstituteValues

Substitute values

(x- 0)^2+(y- 0)^2= 5^2

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Simplify

SubTerms

Subtract terms

x^2+y^2=5^2

CalcPow

Calculate power

x^2+y^2=25

Recall that the x-coordinate of P is 1. Therefore, to find its y-coordinate, this value can be substituted into the equation of the circle.

x^2+y^2=25

Substitute

x= 1

1^2+y^2=25

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Solve for y

BaseOne

1^a=1

1+y^2=25

SubEqn

LHS-1=RHS-1

y^2=24

SqrtEqn

sqrt(LHS)=sqrt(RHS)

y=± sqrt(24)

SplitIntoFactors

Split into factors

y=± sqrt(4(6))

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

y=± sqrt(4)sqrt(6)

CalcRoot

Calculate root

y=± 2sqrt(6)

The y-coordinate of P can be either 2sqrt(6) or - 2sqrt(6). However, from the diagram it can be observed that P is located in Quadrant I, where all values of the y-variable are positive. Therefore, the y-coordinate of P is 2sqrt(6).