# Circle equations # Circle equations on the SAT test

## Studying circle equations

On the SAT test circle equations topic is the last topic of additional topics in math that include 7 advanced topics (see the full topics list on the top menu). It is recommended to start learning additional topics in math with its first topic called complex numbers.

Circle equations topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topic in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math).

### Circle equations- summary

A standard form of a circle contains 2 squares of binomials that are equal to a squared radius, so that a circle with a center (h,k) and a radius r has a standard form equation of (x-h)2+(y-k)2=r2.

To find the location of the circle in the xy plane, we need to draw it according to its standard equation.

If we open brackets of the standard form equation, we get the expended form: x2-2hx+y2-2ky+(h2+k2-r2)=0, in which there are 3 constants h2, k2 and -r2.

Rewriting the expended equation to a standard form: We can identify the first 4 components of the equation since each one of them has a different type of variable (x2,x,y2 and y) but the 3 constants will be given by a single number. We can find the 3 constants by completing 2 quadratic equations of x and y (finding k and h) with square of sum and square of difference formulas.

### A standard form equation of a circle in the xy plane

A standard form of a circle contains 2 squares of binomials that are equal to a squared radius, so that a circle with a center (h,k) and a radius r has a standard form equation of (x-h)2+(y-k)2=r2.

For example: If the center of the circle is located at the point (2,5) and the radius of the circle is equal to 3 then h=a and k=5 the standard equation will be (x-2)2+(y-5)2=32.

The figure below represents a circle in the xy plane and its standard equation.

The center of the circle O has the coordinates of (h,k) and its standard equation is (x-h)2+(y-k)2=r2. ### Drawing a circle in the xy plane using its standard form equation

We can be asked about the location of the circle, like in which quadrant the circle is located. To find the location of the circle in the xy plane, we need to draw it according to its standard equation.

Finding the location of the circle in the xy plane steps:
Step 1: Draw the axes and write the numbers of the quadrants in the xy plane.
Remember that the axes divide the coordinate plane into 4 quadrants, the first quadrant is the top right quadrant (for positive x values and positive y values) and moving counterclockwise. The quadrants are names using the Roman numbers I (+,+), II (-,+) , III (-,-) and IX(+,-).

Step 2: Draw the center of the circle (the center coordinates are h and k values from the equation).

Step 3: Draw the circle by adding the value of the radius around the center (the radius value is written in the equation).

Step 4: See in which quadrant the circle is located.

Consider the following example:

In which quadrant is located the circle that has an equation (x+5)2+(y+5)2=22?

The center of the circle coordinates are (-5,-5) and its radius is equal to 2.

The figure below presents the circle in the xy plane.

The circle is in the third quadrant. ### An expanded equation of a circle

If we open brackets of the standard form equation, we get the expended form:
(x-h)2+(y-k)2=r2
x2+h2-2hx+y2+k2-2ky-r2=0

Note that there are 3 constants in the equations which are h2, k2 and -r2, therefore, we get the equation
x2-2hx+y2-2ky+(h2+k2-r2)=0

### Finding a standard form equation of a circle from its expanded form

To find the center and the radius of the circle, we need to rewrite the expended equation to a standard form.

We can identify the first 4 components of the equation since each one of them has a different type of variable (x2,x,y2 and y) but the 3 constants will be given by a single number. We can find the 3 constants by completing 2 quadratic equations of x and y (finding k and h) with square of sum and square of difference formulas.

We know that the square of sum formula is (a+b)2=a2+b2+2ab.

We know that the square of difference formula is (a-b)2=a2+b2-2ab.

Consider the following example:

We are given the expression x2-6x.

What constant c should be added to this expression, so that we get a quadratic binomial in a form of (x-c)2?

We know that the square of difference formula is (a-b)2=a2+b2-2ab and we are given that a2=x2 and -2ab=-6x

a2=x2

x=a, therefore the first number is x

-2ab=-6x

We found that x=a therefore -2xb=-6x.

We can divide by -2x getting b=-6x/-2x.

b=3, therefore the second number is 3 and the quadratic binomial is (x-3)2

b2=9, therefore the constant that is added to the expression x2-6x to complete a quadratic binomial is 9.

(x-3)2=x2-6x+9

Consider the following example:

We are given the expression 2x2+20x,

What constant c should be added to this expression, so that we get a quadratic binomial in a form of (x-c)2?

First, we should factor out the number 2 to make the coefficient of x2 equal 1, so that the expression will be 2(x2+10x).

We know that the square of sum formula is (a+b)2=a2+b2+2ab and we are given that a2=x2 and 2ab=10x

a2=x2

a=x, therefore the first number is x

2ab=10x

We found that x=a, therefore 2bx=10x

b=5, therefore the second number is 5 and the quadratic binomial is 2(x+5)2

b2=25, therefore the constant that is added to the expression 2x2+20x to complete a quadratic binomial is 25*2=50.

2(x+5)2=2(x2+10x+25)

2(x+5)2=2x2+20x+50

Consider the following example:

What is the center and the radius of the circle equation x2-2x+y2-6y+1=0?

The circle standard equation is (x-h)2+(y-k)2=r2, where the center of the circle is (h,k) and the radius is r.

The given constant of 1 combines 3 constants so that c12+c22-c32=1
c1 for completing the expression x2-2x to quadratic binomial
c2 for completing the expression y2-6y to quadratic binomial
-c3 that is equal to the radius

Note that the sign of c3 is negative since the radius is transferred from the second side of the equation.

We need to complete x2-2x to x2-2x+c12=(x-c1)2 getting x2-2x+12=(x-1)2, therefore c1=1 and c12=1.

We need to complete y2-6y to y2-6y+c22=(x-c2)2 getting y2-6y+32=(y-3)2, therefore c2=3 and c22=9.

We know that c12+c22-c32=1, therefore 1+9-c32=1 and c32=9, c3=3.

x2-2x+1  +  y2-6y+9    +1-10  =0

(x-1)2          (y-3)2          -9     =0

(x-1)2          (y-3)2                  =32

The standard equation is (x-1)2+(y-3)2=32, therefore the center of the circle is (1,3) and the radius of the circle is 3.

(x-1)2+(y-3)2=32

x2+1-2x+y2+9-6y-9=0

x2-2x+y2-6y+1=0

Consider the following example:

What is the center and the radius of the circle equation 4x2-4x+9y2-12y+1=0?

4x2-4x+1  +  9y2-12y+9  +1-10  =0

(2x-1)2           (3y-3)2        -9     =0

(2x-1)2           (3y-3)2                =32

The standard equation is (2x-1)2+(3y-3)2=32, therefore the center of the circle is (1,3) and the radius of the circle is 3.

You just finished studying circle equations topic, the last topic of additional topics in math!