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 $r^2$.
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:
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| 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 |
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[!WARNING] The Two Most Common Circle Traps on the SAT:
- The Sign Reversal Trap: In $(x + 5)^2 + (y - 2)^2 = 36$, the center is $(-5, 2)$, NOT $(5, -2)$.
- 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:
To find the center $(h, k)$ and radius $r$, execute the standardized completing-the-square workflow:
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| 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 |
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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:
- Divide the entire equation by the leading coefficient ($2$):
- Group terms and move constant to the right side:
- Complete the square for $x$ and $y$:
- $x$-constant: $\left(\frac{-8}{2}\right)^2 = (-4)^2 = 16$
- $y$-constant: $\left(\frac{12}{2}\right)^2 = (6)^2 = 36$
- Add constants to BOTH sides:
- Write in standard form:
3. Geometric Properties of Circles in Coordinate Geometry
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| 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) |
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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.
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:
- Find the slope of the radius connecting $C$ and $P$:
- Take the negative reciprocal for the tangent line:
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 $r$.
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 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?
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?