Equations of Circles

Introduction

Circle equations appear in the Geometry and Trigonometry domain, which makes up roughly 15% of the SAT Math section. Questions ask you to write the equation of a circle from given information, rewrite a messy quadratic by completing the square to reveal the center and radius, determine where a specific point sits relative to a circle, and track how edits to the equation shift or resize the graph. All of these — and only these — are covered here. For arc length, sector area, and circle theorems involving tangents and chords, see the sibling notes in this topic.

Core Concept

Every circle in the xy-plane is the set of all points that are exactly $r$ units from a fixed center $(h, k)$ . That distance condition, written using the distance formula (which is just the Pythagorean theorem), gives the center-radius form:

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

Reading the equation: The center is $(h, k)$ and the radius is $r = \sqrt{r^2}$ . The signs inside the parentheses flip: $(x - 3)^2$ means $h = 3$ ; $(y + 2)^2 = (y - (-2))^2$ means $k = -2$ .

Testing a point $(a, b)$ : Substitute into the left side:

Result $= r^2$ : the point is on the circle.
Result $< r^2$ : the point is inside the circle.
Result $> r^2$ : the point is outside the circle.

Shifting a circle: Replacing $x$ with $x - c$ shifts the circle right by $c$ units; replacing $y$ with $y - d$ shifts it up by $d$ units. Equivalently, the center moves from $(h, k)$ to $(h + c, k + d)$ while $r$ stays the same.

Completing the square: The SAT often presents a circle in the expanded form $x^2 + Bx + y^2 + Dy = E$ . To convert it, complete the square for both variables:

x^2 + Bx = \left(x + \tfrac{B}{2}\right)^2 - \tfrac{B^2}{4}

Add the same constant to both sides to keep the equation balanced, then read off $h$ , $k$ , and $r^2$ .

Key Formulas & Rules

Formula	Source
$(x - h)^2 + (y - k)^2 = r^2$ (center-radius form)	Memorize
Distance between $(x_1,y_1)$ and $(x_2,y_2)$ : $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$	Derived from Pythagorean theorem (reference sheet)
Completing the square: $x^2+Bx = \left(x+\frac{B}{2}\right)^2-\frac{B^2}{4}$	Memorize
Point test: substitute $(a,b)$ ; compare result to $r^2$	Memorize

The Pythagorean theorem $c^2 = a^2 + b^2$ is on the reference sheet; the distance formula and circle equation must be memorized.

Worked Examples

Example 1

A circle in the xy-plane has center $(3, -2)$ and radius $5$ . Which of the following is an equation of the circle?

A) $(x + 3)^2 + (y - 2)^2 = 5$

B) $(x - 3)^2 + (y + 2)^2 = 25$

C) $(x - 3)^2 + (y - 2)^2 = 25$

D) $(x + 3)^2 + (y + 2)^2 = 25$

Solution:

The center is $(h, k) = (3, -2)$ and $r = 5$ , so $r^2 = 25$ .

Plug into the center-radius form:

(x - 3)^2 + (y - (-2))^2 = 25 \implies (x-3)^2 + (y+2)^2 = 25

Why the distractors fail:

A: Both signs are flipped ( $+3$ instead of $-3$ , $-2$ instead of $+2$ ) and the right side shows $r = 5$ instead of $r^2 = 25$ .
C: The $x$ -term is correct, but $(y - 2)^2$ uses $k = 2$ instead of $k = -2$ — a sign error on $k$ .
D: Only the $x$ -term sign is wrong: $(x + 3)^2$ encodes $h = -3$ instead of $h = 3$ , giving center $(-3, -2)$ instead of $(3, -2)$ .

Answer: B

Example 2

The equation $x^2 + 8x + y^2 - 2y = 3$ represents a circle in the xy-plane. What are the center and radius of this circle?

A) Center $(-4, 1)$ , radius $2\sqrt{5}$

B) Center $(4, -1)$ , radius $2\sqrt{5}$

C) Center $(-4, 1)$ , radius $\sqrt{3}$

D) Center $(-4, 1)$ , radius $20$

Solution:

Complete the square for $x$ and $y$ separately.

For $x$ : Coefficient of $x$ is $8$ ; half is $4$ ; square is $16$ .

x^2 + 8x = (x + 4)^2 - 16

For $y$ : Coefficient of $y$ is $-2$ ; half is $-1$ ; square is $1$ .

y^2 - 2y = (y - 1)^2 - 1

Substitute back and add the constants to both sides:

(x+4)^2 - 16 + (y-1)^2 - 1 = 3

(x+4)^2 + (y-1)^2 = 3 + 16 + 1 = 20

Center: $(-4, 1)$ ; $r^2 = 20$ ; $r = \sqrt{20} = 2\sqrt{5}$ .

Why the distractors fail:

B: Flips the signs on $h$ and $k$ to get center $(4,-1)$ — a common "forget to flip" error.
C: Uses the original right-hand side $3$ as $r^2$ without adding the completing-the-square constants $16$ and $1$ .
D: Reads $r^2 = 20$ as $r = 20$ rather than taking the square root.

Desmos check

Desmos check: Enter x^2+8x+y^2-2y=3 into Desmos. The graph shows a circle; clicking on it reveals the center and radius, confirming $(-4,1)$ and $2\sqrt{5}$ .

Answer: A

Common Mistakes & Traps

Sign flip on center. $(x + 4)^2 + (y - 1)^2 = r^2$ has center $(-4, 1)$ , not $(4, 1)$ or $(4, -1)$ . Always rewrite as $(x - h)^2$ and read $h$ directly.
Forgetting to add completing-the-square constants to the right side. When you add $\left(\frac{B}{2}\right)^2$ to the left side, you must add the same amount to the right. Missing this step produces a wrong $r^2$ .
Confusing $r^2$ with $r$ . The equation gives $r^2$ directly. Take the square root to find the radius. SAT options for radius and $r^2$ are both usually listed as traps.
Shifting in the wrong direction. Shifting the circle right by $c$ changes center $(h,k)$ to $(h+c, k)$ , which changes $(x-h)^2$ to $(x-(h+c))^2$ . Do not add $c$ directly to the expression inside the parentheses.
Misidentifying "inside" vs "on." If the point test gives exactly $r^2$ , the point is on the circle — not inside. The SAT specifically designs distractors using points on the circle when you are asked for a point strictly inside.

Practice Questions

Question 1

A circle is represented by $(x - 2)^2 + (y + 3)^2 = 16$ . The circle is shifted $3$ units to the right and $5$ units up. Which of the following is the equation of the resulting circle?

A) $(x - 5)^2 + (y - 2)^2 = 16$

B) $(x + 1)^2 + (y + 8)^2 = 16$

C) $(x - 5)^2 + (y + 8)^2 = 16$

D) $(x - 5)^2 + (y - 2)^2 = 21$

Show answer

Answer: A

The original center is $(h, k) = (2, -3)$ . Shifting right $3$ and up $5$ :

(2 + 3,\; -3 + 5) = (5,\; 2)

The radius (and therefore $r^2 = 16$ ) is unchanged.

New equation: $(x - 5)^2 + (y - 2)^2 = 16$ .

Why the others fail:

B: Adds the shift amounts directly inside the expressions: $(x - 2 + 3) = (x + 1)$ and $(y + 3 + 5) = (y + 8)$ , which would encode center $(-1, -8)$ — this misunderstands how the equation encodes the center.
C: Shifts $x$ correctly (giving $h = 5$ ), but adds the $5$ -unit shift directly inside the $y$ -expression: $(y + 3 + 5) = (y + 8)$ , encoding $k = -8$ . The net effect on the center's $y$ -coordinate is a move from $-3$ to $-8$ , i.e., down $5$ instead of up $5$ .
D: Correctly updates the center to $(5, 2)$ but then adds the vertical shift amount to $r^2$ : $16 + 5 = 21$ . Translations never change the radius.

Question 2

A circle in the xy-plane has equation $(x - 3)^2 + (y + 2)^2 = 49$ . Which of the following points lies strictly inside the circle?

A) $(10, -2)$

B) $(3, 5)$

C) $(8, 2)$

D) $(-5, -2)$

Show answer

Answer: C

Substitute each point into $(x-3)^2+(y+2)^2$ and compare with $49$ :

Point	Calculation	Value	Position
$(10,-2)$	$7^2 + 0^2$	$49$	On
$(3,5)$	$0^2 + 7^2$	$49$	On
$(8,2)$	$5^2 + 4^2$	$41 < 49$	Inside
$(-5,-2)$	$(-8)^2 + 0^2$	$64 > 49$	Outside

Options A and B are on the circle (not strictly inside); D is outside. Only C satisfies the condition of lying strictly inside.

Question 3

A circle is represented by $(x + 1)^2 + (y - 3)^2 = 9$ . If the radius of the circle is doubled, which equation represents the new circle?

A) $(x + 1)^2 + (y - 3)^2 = 18$

B) $(x + 1)^2 + (y - 3)^2 = 36$

C) $(x + 1)^2 + (y - 3)^2 = 81$

D) $(2x + 2)^2 + (2y - 6)^2 = 9$

Show answer

Answer: B

The original radius satisfies $r^2 = 9$ , so $r = 3$ . Doubling the radius gives $r_{\text{new}} = 6$ , so $r_{\text{new}}^2 = 36$ .

The center $(-1, 3)$ is unchanged.

New equation: $(x+1)^2+(y-3)^2=36$ .

Why the others fail:

A: Doubles $r^2$ directly: $9 \times 2 = 18$ . But doubling the radius multiplies $r^2$ by $4$ (since $(2r)^2 = 4r^2$ ), not by $2$ .
C: Squares $r^2$ itself: $9^2 = 81$ . This is a "double-squaring" error — confusing $r^2 = 9$ with $r = 9$ and then squaring.
D: Doubles the coefficients inside the squared expressions instead of changing $r^2$ . This does not simply scale the radius; it distorts the equation entirely and does not represent a standard circle in center-radius form.

Question 4 (Student-produced response)

The equation $x^2 - 4x + y^2 + 6y = 12$ represents a circle. What is the radius of this circle?

Show answer

Answer: 5

Complete the square:

x^2 - 4x = (x - 2)^2 - 4

y^2 + 6y = (y + 3)^2 - 9

Substitute:

(x-2)^2 - 4 + (y+3)^2 - 9 = 12

(x-2)^2 + (y+3)^2 = 12 + 4 + 9 = 25

So $r^2 = 25$ and $r = 5$ .

Center: $(2, -3)$ ; Radius: $5$ .

Enter $5$ in the answer box.

Question 5 (Student-produced response)

The equation $x^2 + 8x + y^2 - 2y = 3$ represents a circle in the xy-plane. Point $P$ is the center of this circle, and point $Q$ has coordinates $(0, 0)$ . What is the distance from $P$ to $Q$ ? Give your answer as a simplified radical or an integer.

Show answer

Answer: $\sqrt{17}$

From the worked example in this note, completing the square gives center $P = (-4, 1)$ .

Distance from $P = (-4, 1)$ to $Q = (0, 0)$ :

d = \sqrt{(-4-0)^2+(1-0)^2} = \sqrt{16+1} = \sqrt{17}

Enter $\sqrt{17}$ in the answer box (or the decimal approximation $4.123$ , rounded to three decimal places, if the problem permits a decimal entry).

Connections

Pythagorean theorem and special right triangles (prerequisite): the distance formula and the circle equation are both direct applications of $a^2 + b^2 = c^2$ . Recognizing a $3$ - $4$ - $5$ or $5$ - $12$ - $13$ Pythagorean triple can let you test points or find radii without a calculator.
Arc Length and Sector Area (sibling note): once you know $r$ from the equation, you can compute arc lengths and sector areas. These skills combine on multi-part problems.
Radians and the Unit Circle (sibling note): the unit circle is just the special case $(x-0)^2+(y-0)^2=1$ ; understanding its equation grounds the trigonometric coordinate definitions.
Circle Theorems: Tangents, Chords, and Angles (sibling note): a tangent from an external point to a circle uses the distance formula alongside the circle equation — both skills work together.
Completing the square also appears in converting quadratic equations to vertex form (Advanced Math domain), so mastering it here pays double dividends on test day.

Introduction

Core Concept

Key Formulas & Rules

Worked Examples

Example 1

Example 2

Common Mistakes & Traps

Practice Questions

Connections

Figures

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