**Included topics:**linear functions; linear equations; systems of linear equations; linear inequalities.

**Included topics:**quadratic equations and functions; graphing quadratic functions; linear and quadratic systems of equations; exponential expressions; graphing exponential functions; polynomial functions and graphs; radical and rational equations and expressions; isolating quantities; function notation.

**Included topics:**ratios rates and proportions; percentages; units and unit conversion; linear and exponential growth; table data; data collection and data inference; center spread and shape of distributions; key features of graphs; scatterplots.

**Included topics:**complex numbers; volume word problems; congruence and similarity; right triangle trigonometry and word problems; circle theorems; angles, arc lengths and trig functions; circle equations.

# Circle equations on the SAT test

## SAT Subscore: Additional topics in math

### 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}=r^{2}.

**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**: x^{2}-2hx+y^{2}-2ky+(h^{2}+k^{2}-r^{2})=0, in which there are 3 constants h^{2}, k^{2} and -r^{2}.

**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 (x^{2},x,y^{2} 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.

**Continue reading this page for detailed explanations and examples.**

### 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}=r^{2}.

**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}=3^{2}.

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}=r^{2}.

### 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.

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

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

To learn more about quadratic equations on quadratic equations and quadratic functions page.

If we open brackets of the standard form equation, we get the expended form:

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

x^{2}+h^{2}-2hx+y^{2}+k^{2}-2ky-r^{2}=0

Note that there are** 3 constants** in the equations which are h^{2}, k^{2} and -r^{2}, therefore, we get the equation

x^{2}-2hx+y^{2}-2ky+**(h ^{2}+k^{2}-r^{2})**=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 (x^{2},x,y^{2} 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}=a^{2}+b^{2}+2ab.

We know that the square of difference formula is (a-b)^{2}=a^{2}+b^{2}-2ab.

We are given the expression x^{2}-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}=a^{2}+b^{2}-2ab and we are given that a^{2}=x^{2} and -2ab=-6x

a^{2}=x^{2}

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}

b^{2}=9, therefore the constant that is added to the expression x^{2}-6x to complete a quadratic binomial is 9.

__Checking the answer:__

(x-3)^{2}=x^{2}-6x+9

We are given the expression 2x^{2}+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 x^{2} equal 1, so that the expression will be 2(x^{2}+10x).

We know that the square of sum formula is (a+b)^{2}=a^{2}+b^{2}+2ab and we are given that a^{2}=x^{2} and 2ab=10x

a^{2}=x^{2}

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}

b^{2}=25, therefore the constant that is added to the expression 2x^{2}+20x to complete a quadratic binomial is 25*2=50.

__Checking the answer:__

2(x+5)^{2}=2(x^{2}+10x+25)

2(x+5)^{2}=2x^{2}+20x+50

What is the center and the radius of the circle equation x^{2}-2x+y^{2}-6y+1=0?

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

**The given constant of 1** combines 3 constants so that c_{1}^{2}+c_{2}^{2}-c_{3}^{2}=1

c_{1} for completing the expression x^{2}-2x to quadratic binomial

c_{2} for completing the expression y^{2}-6y to quadratic binomial

-c_{3} that is equal to the radius

**Note that the sign of c _{3} is negative** since the radius is transferred from the second side of the equation.

We need to complete x^{2}-2x to x^{2}-2x+c_{1}^{2}=(x-c_{1})^{2} getting x^{2}-2x+1^{2}=(x-1)^{2}, therefore c_{1}=1 and c_{1}^{2}=1.

We need to complete y^{2}-6y to y^{2}-6y+c_{2}^{2}=(x-c_{2})^{2} getting y^{2}-6y+3^{2}=(y-3)^{2}, therefore c_{2}=3 and c_{2}^{2}=9.

We know that c_{1}^{2}+c_{2}^{2}-c_{3}^{2}=1, therefore 1+9-c_{3}^{2}=1 and c_{3}^{2}=9, c_{3}=3.

x^{2}-2x+1 + y^{2}-6y+9 +1-10 =0

(x-1)^{2} (y-3)^{2} -9 =0

(x-1)^{2} (y-3)^{2} =3^{2}

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

Checking the answer:

(x-1)^{2}+(y-3)^{2}=3^{2}

x^{2}+1-2x+y^{2}+9-6y-9=0

x^{2}-2x+y^{2}-6y+1=0

What is the center and the radius of the circle equation 4x^{2}-4x+9y^{2}-12y+1=0?

4x^{2}-4x+1 + 9y^{2}-12y+9 +1-10 =0

(2x-1)^{2} (3y-3)^{2} -9 =0

(2x-1)^{2} (3y-3)^{2} =3^{2}

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