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:
x2 + y2 + 6x – 4y = –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 x2 + 6x tells you that (x + 3)2 is part of the equation. FOIL this out to x2 + 6x + 9. However, the x2 + 6x is by itself on the left, so add 9 to both sides of the equation:
Also, y2 – 4y tells you that (y – 2)2 is part of the equation, which FOILs out to y2 – 4y + 4. However, the y2 – 4y is by itself on the left, so add 4 to both sides, like this:
To convert the circle to its standard form, factor x2 + 6x + 9 into (x + 3)2 and y2 – 4y + 4 into (y – 2)2, like this:
Now the circle is in its familiar form, and r2 = 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 x2 + 6x tells you that (x + 3)2 is part of the equation. FOIL this out to x2 + 6x + 9. However, the x2 + 6x is by itself on the left, so add 9 to both sides of the equation:
Also, y2 – 4y tells you that (y – 2)2 is part of the equation, which FOILs out to y2 – 4y + 4. However, the y2 – 4y is by itself on the left, so add 4 to both sides, like this:
To convert the circle to its standard form, factor x2 + 6x + 9 into (x + 3)2 and y2 – 4y + 4 into
(y – 2)2, like this:
Now the circle is in its familiar form, where h = –3 and k = 2.
