# SAT Math Problems: Using the Standard Circle Equation

If you encounter a question on the SAT Math exam that gives you the equation of a circle, you’ll probably need to convert that equation to the standard circle equation.

The following practice questions give you the equation of a circle and ask you to find its radius and center.

## Practice questions

Questions 1 and 2 are based on the following information.

The equation of a circle in the *xy*-plane is shown here:

*x*^{2} + *y*^{2} + 6*x* – 4*y* = –9

**What is the radius of the circle?**

**A.**1

**B.**2

**C.**3

**D.**4**What are the (***x*,*y*) coordinates of the center?

**A.**(–3, 2)

**B.**(–2, 3)

**C.**(3, –2)

**D.**(2, –3)

## Answers and explanations

**The correct answer is Choice (B).**

First convert the equation to the standard circle equation:

where*r*is the radius of the circle. From the original equation, start by moving the*x*‘s and*y*‘s together:

The*x*^{2}+ 6*x*tells you that (*x*+ 3)^{2}is part of the equation. FOIL this out to*x*^{2}+ 6*x*+ 9. However, the*x*^{2}+ 6*x*is by itself on the left, so add 9 to both sides of the equation:

Also,*y*^{2}– 4*y*tells you that (*y*– 2)^{2}is part of the equation, which FOILs out to*y*^{2}– 4*y*+ 4. However, the*y*^{2}– 4*y*is by itself on the left, so add 4 to both sides, like this:

To convert the circle to its standard form, factor*x*^{2}+ 6*x*+ 9 into (*x*+ 3)^{2}and*y*^{2}– 4*y*+ 4 into (*y*– 2)^{2}, like this:

Now the circle is in its familiar form, and*r*^{2}= 4, so*r*= 2.**The correct answer is Choice (A).**

First convert the equation to the standard circle equation:

where*h*is the*x*-coordinate and*k*is the*y*-coordinate of the center of the circle. From the original equation, start by moving the*x*‘s and*y*‘s together:

The*x*^{2}+ 6*x*tells you that (*x*+ 3)^{2}is part of the equation. FOIL this out to*x*^{2}+ 6*x*+ 9. However, the*x*^{2}+ 6*x*is by itself on the left, so add 9 to both sides of the equation:

Also,*y*^{2}– 4*y*tells you that (*y*– 2)^{2}is part of the equation, which FOILs out to*y*^{2}– 4*y*+ 4. However, the*y*^{2}– 4*y*is by itself on the left, so add 4 to both sides, like this:

To convert the circle to its standard form, factor*x*^{2}+ 6*x*+ 9 into (*x*+ 3)^{2}and*y*^{2}– 4*y*+ 4 into

(*y*– 2)^{2}, like this:

Now the circle is in its familiar form, where*h*= –3 and*k*= 2.