7.5 Circle Equations, Center & Radius in Coordinate Plane

Key Takeaways

  • The standard form of a circle equation is $(x - h)^2 + (y - k)^2 = r^2$, with center $(h, k)$ and radius $r = \sqrt{r^2}$; always reverse the signs inside the parentheses.
  • To convert the general conic form $x^2 + y^2 + Ax + By + C = 0$ to standard form, complete the square separately for $x$ and $y$ by adding $(\frac{A}{2})^2$ and $(\frac{B}{2})^2$ to both sides.
  • The line tangent to a circle is perpendicular to the radius at the point of tangency, meaning their slopes are negative reciprocals: $m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}}$.
  • If diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ are given, the center $(h, k)$ is their midpoint and the radius $r$ is half the distance between them.
  • A point $(x_0, y_0)$ lies inside the circle if $(x_0 - h)^2 + (y_0 - k)^2 < r^2$, on the circle if equal to $r^2$, and outside the circle if greater than $r^2$.
Last updated: August 2026

7.5 Circle Equations, Center & Radius in Coordinate Plane

Circle equations in the $xy$-plane are guaranteed to appear on the Digital SAT Math section (typically 1 to 2 questions per test). These questions test two primary skills: extracting the center and radius from the standard form, and converting a general quadratic equation into standard form by completing the square.


1. Standard Form of a Circle Equation

A circle is the set of all points $(x, y)$ equidistant from a fixed center point $(h, k)$. Applying the Pythagorean distance formula yields the standard form:

(xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2

+-----------------------------------------------------------------------------+
|                        STANDARD CIRCLE EQUATION                             |
|                                                                             |
|                           (x - h)² + (y - k)² = r²                          |
|                                                                             |
|   - Center Coordinates:   (h, k)  <--- (NOTE THE SIGN REVERSAL!)            |
|   - Radius Length:        r = sqrt(r²)                                      |
|                                                                             |
|   [ EXAMPLES ]:                                                             |
|   1. (x - 3)² + (y - 5)² = 49    ===>  Center: (3, 5),   Radius: sqrt(49)=7 |
|   2. (x + 4)² + (y - 1)² = 20    ===>  Center: (-4, 1),  Radius: sqrt(20)   |
|   3. x² + (y + 6)² = 16          ===>  Center: (0, -6),  Radius: sqrt(16)=4 |
+-----------------------------------------------------------------------------+

[!WARNING] The Two Most Common Circle Traps on the SAT:

  1. The Sign Reversal Trap: In $(x + 5)^2 + (y - 2)^2 = 36$, the center is $(-5, 2)$, NOT $(5, -2)$.
  2. The $r^2$ Trap: The number on the right side is $r^2$, NOT $r$. If the equation equals $36$, the radius is $\sqrt{36} = 6$, NOT $36$.

2. Converting General Form via Completing the Square

SAT questions frequently present circle equations in expanded general form:

x2+y2+Ax+By+C=0x^2 + y^2 + Ax + By + C = 0

To find the center $(h, k)$ and radius $r$, execute the standardized completing-the-square workflow:

+-----------------------------------------------------------------------------+
|                   COMPLETING THE SQUARE 5-STEP WORKFLOW                     |
|                                                                             |
|   Given:  x² + y² - 10x + 6y + 9 = 0                                        |
|                                                                             |
|   [STEP 1: REARRANGE & GROUP TERMS]                                         |
|   Group x-terms and y-terms; move constant to the right side:               |
|   (x² - 10x) + (y² + 6y) = -9                                               |
|                                                                             |
|   [STEP 2: FIND SQUARING CONSTANTS]                                         |
|   Take half of each linear coefficient and square it:                       |
|   For x:  (-10 / 2)² = (-5)² = 25                                           |
|   For y:  (6 / 2)²   = (3)²  = 9                                            |
|                                                                             |
|   [STEP 3: ADD CONSTANTS TO BOTH SIDES]                                     |
|   (x² - 10x + 25) + (y² + 6y + 9) = -9 + 25 + 9                             |
|                                                                             |
|   [STEP 4: FACTOR INTO PERFECT SQUARES]                                     |
|   (x - 5)² + (y + 3)² = 25                                                  |
|                                                                             |
|   [STEP 5: IDENTIFY CENTER AND RADIUS]                                      |
|   Center = (5, -3),   Radius r = sqrt(25) = 5                               |
+-----------------------------------------------------------------------------+

Worked Example 1: Completing the Square with Leading Coefficients

Problem: The equation of a circle is given by $2x^2 + 2y^2 - 16x + 24y - 18 = 0$. What is the radius of the circle?

Step-by-Step Solution:

  1. Divide the entire equation by the leading coefficient ($2$): x2+y28x+12y9=0x^2 + y^2 - 8x + 12y - 9 = 0
  2. Group terms and move constant to the right side: (x28x)+(y2+12y)=9(x^2 - 8x) + (y^2 + 12y) = 9
  3. Complete the square for $x$ and $y$:
    • $x$-constant: $\left(\frac{-8}{2}\right)^2 = (-4)^2 = 16$
    • $y$-constant: $\left(\frac{12}{2}\right)^2 = (6)^2 = 36$
  4. Add constants to BOTH sides: (x28x+16)+(y2+12y+36)=9+16+36=61(x^2 - 8x + 16) + (y^2 + 12y + 36) = 9 + 16 + 36 = 61
  5. Write in standard form: (x4)2+(y+6)2=61(x - 4)^2 + (y + 6)^2 = 61 Radius r=61\text{Radius } r = \sqrt{61}

3. Geometric Properties of Circles in Coordinate Geometry

+-----------------------------------------------------------------------------+
|                      COORDINATE GEOMETRY CIRCLE THEOREMS                    |
|                                                                             |
|   [ TANGENT LINE PERPENDICULARITY ]     [ DIAMETER MIDPOINT & RADIUS ]      |
|                                                                             |
|                   \  Tangent Line                    Point A (x1, y1)       |
|                    \  (slope m_t)                     /                     |
|                     \                                /                      |
|             Center   \ P (tangency)                 /                       |
|             (h, k)----+                            + Center (Midpoint)      |
|               \      /                            /                         |
|                \    /                            /                          |
|           Radius (m_r)                          /                           |
|                                             Point B (x2, y2)                |
|   Perpendicular Slopes:                                                     |
|   m_tangent = -1 / m_radius             Center = ((x1+x2)/2, (y1+y2)/2)     |
|                                         Radius = 1/2 * Distance(A, B)       |
+-----------------------------------------------------------------------------+

1. Tangent Line Perpendicularity

A line tangent to a circle intersects the circle at exactly one point (the point of tangency). The tangent line is perpendicular to the radius drawn to that point.

mtangent=1mradiusm_{\text{tangent}} = -\frac{1}{m_{\text{radius}}}

Worked Example 2: Finding the Slope of a Tangent Line

Problem: A circle has center $C(1, -3)$ and passes through point $P(4, 1)$. A line is tangent to the circle at point $P$. What is the slope of the tangent line?

Step-by-Step Solution:

  1. Find the slope of the radius connecting $C$ and $P$: mradius=y2y1x2x1=1(3)41=43m_{\text{radius}} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-3)}{4 - 1} = \frac{4}{3}
  2. Take the negative reciprocal for the tangent line: mtangent=14/3=34m_{\text{tangent}} = -\frac{1}{4/3} = -\frac{3}{4}

4. Point Location Relative to a Circle

To determine whether a given point $(x_0, y_0)$ lies inside, on, or outside a circle with center $(h, k)$ and radius $r$, compute the squared distance $D^2 = (x_0 - h)^2 + (y_0 - k)^2$:

  • Inside the Circle: $(x_0 - h)^2 + (y_0 - k)^2 < r^2$
  • On the Boundary: $(x_0 - h)^2 + (y_0 - k)^2 = r^2$
  • Outside the Circle: $(x_0 - h)^2 + (y_0 - k)^2 > r^2$

[!TIP] Desmos Instant Circle Verification: On the Digital SAT, type the general circle equation directly into Desmos (e.g., x^2 + y^2 + 8x - 12y - 12 = 0). Desmos plots the circle automatically. Click the leftmost and rightmost points to find the center $(h, k)$ as the horizontal midpoint and count grid units to find the radius $r$.

Test Your Knowledge

The equation of a circle in the xy-plane is given by x² + y² + 8x - 12y - 12 = 0. What are the coordinates of the center (h, k) and the radius r of this circle?

A
B
C
D
Test Your Knowledge

A circle in the xy-plane is defined by (x - 3)² + (y + 2)² = 25. A line is tangent to this circle at the point (6, 2). What is the slope of the tangent line?

A
B
C
D
Test Your Knowledge

The endpoints of a diameter of a circle in the xy-plane are (-1, 5) and (7, -1). Which of the following equations represents this circle?

A
B
C
D