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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=r2(x - h)^2 + (y - k)^2 = r^2, with center (h,k)(h, k) and radius r=r2r = \sqrt{r^2}; always reverse the signs inside the parentheses.

  • To convert the general conic form x2+y2+Ax+By+C=0x^2 + y^2 + Ax + By + C = 0 to standard form, complete the square separately for xx and yy by adding (A2)2(\frac{A}{2})^2 and (B2)2(\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: mtangent=−1mradiusm_{\text{tangent}} = -\frac{1}{m_{\text{radius}}}.

  • If diameter endpoints (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are given, the center (h,k)(h, k) is their midpoint and the radius rr is half the distance between them.

  • A point (x0,y0)(x_0, y_0) lies inside the circle if (x0−h)2+(y0−k)2<r2(x_0 - h)^2 + (y_0 - k)^2 < r^2, on the circle if equal to r2r^2, and outside the circle if greater than r2r^2.

Last updated: August 2026

7.5 Circle Equations, Center & Radius in Coordinate Plane

Circle equations in the xyxy-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)(x, y) equidistant from a fixed center point (h,k)(h, k). Applying the Pythagorean distance formula yields the standard form:

(x−h)2+(y−k)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(x + 5)^2 + (y - 2)^2 = 36, the center is (−5,2)(-5, 2), NOT (5,−2)(5, -2).
  2. The r2r^2 Trap: The number on the right side is r2r^2, NOT rr. If the equation equals 3636, the radius is 36=6\sqrt{36} = 6, NOT 3636.

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)(h, k) and radius rr, 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 2x2+2y2−16x+24y−18=02x^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 (22): x2+y2−8x+12y−9=0x^2 + y^2 - 8x + 12y - 9 = 0
  2. Group terms and move constant to the right side: (x2−8x)+(y2+12y)=9(x^2 - 8x) + (y^2 + 12y) = 9
  3. Complete the square for xx and yy:
    • xx-constant: (−82)2=(−4)2=16\left(\frac{-8}{2}\right)^2 = (-4)^2 = 16
    • yy-constant: (122)2=(6)2=36\left(\frac{12}{2}\right)^2 = (6)^2 = 36
  4. Add constants to BOTH sides: (x2−8x+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: (x−4)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)C(1, -3) and passes through point P(4,1)P(4, 1). A line is tangent to the circle at point PP. What is the slope of the tangent line?

Step-by-Step Solution:

  1. Find the slope of the radius connecting CC and PP: mradius=y2−y1x2−x1=1−(−3)4−1=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 (x0,y0)(x_0, y_0) lies inside, on, or outside a circle with center (h,k)(h, k) and radius rr, compute the squared distance D2=(x0−h)2+(y0−k)2D^2 = (x_0 - h)^2 + (y_0 - k)^2:

  • Inside the Circle: (x0−h)2+(y0−k)2<r2(x_0 - h)^2 + (y_0 - k)^2 < r^2
  • On the Boundary: (x0−h)2+(y0−k)2=r2(x_0 - h)^2 + (y_0 - k)^2 = r^2
  • Outside the Circle: (x0−h)2+(y0−k)2>r2(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)(h, k) as the horizontal midpoint and count grid units to find the radius rr.

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

Center (4, -6) and radius r = 8

B

Center (-4, 6) and radius r = 8

C

Center (-4, 6) and radius r = 64

D

Center (8, -12) and radius r = 16

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

-3/4

B

-4/3

C

3/4

D

4/3

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

(x - 4)² + (y - 3)² = 100

B

(x - 3)² + (y - 2)² = 100

C

(x - 3)² + (y - 2)² = 25

D

(x + 3)² + (y + 2)² = 25

Sections you finish are checked off in the contents.

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7.6 Arc Length, Sector Area & 3D Geometry / Volume

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