Circle Equations: Center and Radius
The equation $(x - h)^2 + (y - k)^2 = r^2$ tells you a circle's center $(h, k)$ and radius $r$ at a glance — if you read the signs carefully.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
The center is $(h, k)$, but note the minus signs: $(x + 3)^2$ means $h = -3$. The right side is $r^2$, so take the square root for the radius.
How to solve one, step by step
Example: $(x - 3)^2 + (y + 2)^2 = 25$.
- Center: $(3, -2)$ (flip the sign inside each).
- Radius: $\sqrt{25} = 5$.
The mistakes that cost points
- Not negating to find the center. $(x + 3)^2$ gives $h = -3$, not $+3$.
- Forgetting to take the square root. The right side is $r^2$; the radius is its square root.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
A circle has equation $(x + k)^2 + (y - 2)^2 = 49$. The center of the circle has $x$-coordinate $-4$. What is the value of $k$?
- A$-4$
- B$2$
- C$7$
- D$4$
Show solution
Answer: D, $4$. The term $(x + k)^2 = (x - (-k))^2$, so the center's $x$-coordinate is $-k$. Setting $-k = -4$ gives $k = 4$.
Medium
A circle with equation $(x-2)^2 + (y+2)^2 = 25$ is moved $5$ units right and $1$ unit down. The radius of the circle is increased by $1$ unit. What is the equation of this new circle?
- A$(x+3)^2 + (y+1)^2 = 26$
- B$(x-7)^2 + (y+3)^2 = 36$
- C$(x-7)^2 + (y+3)^2 = 26$
- D$(x+3)^2 + (y+1)^2 = 36$
Show solution
Answer: B, $(x-7)^2 + (y+3)^2 = 36$. The original circle has centre $(2, -2)$ and radius $5$. Moving the circle $5$ units right and $1$ unit down gives a new centre of $(7, -3)$. The radius is increased by $1$ unit, giving a new radius of $6$, so the right-hand side of the equation becomes $6^2 = 36$. The new equation is $(x-7)^2 + (y+3)^2 = 36$.
Hard
A circle has equation $(x - k)^2 + (y + 1)^2 = k^2 + 9$. The circle passes through the point $(k + 3, -1)$. What is the value of $k$?
- A$3$
- B$-1$
- C$0$
- D$9$
Show solution
Answer: C, $0$. The center is $(k, -1)$. Since the point $(k+3, -1)$ is on the circle, the distance from center to point equals $r$. Distance $= \sqrt{(k+3-k)^2 + 0} = 3$. So $r = 3$, meaning $r^2 = 9$. But $r^2 = k^2 + 9$, so $k^2 + 9 = 9$, giving $k^2 = 0$ and $k = 0$.
Find your exact gaps
Take a free, full-length practice test and see precisely which question types trip you up.
Practice this for freeNo credit card. Register once, take as many tests as you like.