Geometry Trigonometry⚡ High Yield (1-3 Questions per Test)
Circle Equations
Digital SAT Math Preparation & Desmos Strategies
4 Concepts18 Practice Qs30 Mock Qs⚡ Desmos Speed Hacks
Key Concepts & Worked Archetypes
Concept 1
Concept 1: Standard Form of a Circle Equation
A circle is defined as the set of all points in a plane that are equidistant from a fixed center point $(h, k)$ by a constant radius $r$.
Standard equation: $(x - h)^2 + (y - k)^2 = r^2$
Center coordinates are given by $(h, k)$, noting the sign reversal inside the binomials.
Radius is the square root of the constant term on the right side: $r = \sqrt{r^2}$.
📘 Traditional Algebraic Method
Identify the center coordinates by taking the opposite sign of the constants inside the parentheses. Take the square root of the right-hand side constant to find the radius length.
⚡ SAT Speed Trick & Desmos Hack
Type the given circle equation directly into the Desmos graphing calculator. Click on the center point and the x/y intercepts to instantly read off coordinates and lengths.
💡 Worked SAT Archetype Example
Problem: What is the radius of the circle given by the equation $(x + 3)^2 + (y - 5)^2 = 49$?
📘 Step-by-Step Textbook Solution:
Step 1
Identify the standard equation form: $(x - h)^2 + (y - k)^2 = r^2$.
Step 2
Match the right side of the given equation to $r^2$: $r^2 = 49$.
Step 3
Solve for the radius $r$ by taking the principal square root: $r = \sqrt{49}$.
Step 4
State the final calculated value: $r = 7$.
⚡ Speed / Desmos Tactic:
Step 1
Open the Desmos graphing calculator app.
Step 2
Type $(x+3)^2 + (y-5)^2 = 49$ into expression line 1.
Step 3
Click on the plotted circle to view its center and extrema, yielding radius = 7 instantly.
Concept 2
Concept 2: Completing the Square for General Form
Circle equations are frequently presented in an expanded general polynomial form that requires algebraic rearrangement to reveal the center and radius.
General form: $x^2 + y^2 + Dx + Ey + F = 0$
Group $x$ terms and $y$ terms together, moving the constant $F$ to the right side.
Add $(\frac{D}{2})^2$ and $(\frac{E}{2})^2$ to both sides of the equation to complete the squares.
📘 Traditional Algebraic Method
Group $x$ and $y$ variables, factor out coefficients if needed, complete the square for both binomials, balance both sides, and rewrite in standard factored form.
⚡ SAT Speed Trick & Desmos Hack
Avoid manual algebra; type the general form equation directly into Desmos. Click the center point and use distance formula or read the grid lines to determine the radius.
💡 Worked SAT Archetype Example
Problem: Find the radius of the circle defined by $x^2 + y^2 - 6x + 4y - 12 = 0$.
📘 Step-by-Step Textbook Solution:
Step 1
Group $x$ and $y$ terms together: $(x^2 - 6x) + (y^2 + 4y) = 12$.
Step 2
Complete the square for $x$: add $(\frac{-6}{2})^2 = 9$ to both sides.
Step 3
Complete the square for $y$: add $(\frac{4}{2})^2 = 4$ to both sides.
Step 4
Rewrite equation: $(x - 3)^2 + (y + 2)^2 = 25$.
Step 5
Determine radius: $r = \sqrt{25} = 5$.
⚡ Speed / Desmos Tactic:
Step 1
Input $x^2 + y^2 - 6x + 4y - 12 = 0$ into Desmos.
Step 2
Click the highest point or rightmost point on the circle to measure distance from the center $(3, -2)$.
Step 3
Read radius = 5 directly.
Concept 3
Concept 3: Finding Center Coordinates from Expanded Forms
Questions often ask specifically for the x-coordinate or y-coordinate of a circle's center given in expanded or semi-factored form.
For any expanded $x$ component of the form $x^2 + Dx$, the center coordinate $h$ is $-\frac{D}{2}$.
For any expanded $y$ component of the form $y^2 + Ey$, the center coordinate $k$ is $-\frac{E}{2}$.
Always verify signs carefully to avoid common sign-reversal errors.
📘 Traditional Algebraic Method
Apply the vertex/center formulas $h = -D/2$ and $k = -E/2$, or fully complete the square for both variables to isolate the center coordinates $(h, k)$.
⚡ SAT Speed Trick & Desmos Hack
Use Desmos to graph the equation, click on the center point label that appears automatically upon hovering over the middle of the circle.
💡 Worked SAT Archetype Example
Problem: What is the y-coordinate of the center of the circle represented by $x^2 + y^2 + 10x - 8y + 5 = 0$?
📘 Step-by-Step Textbook Solution:
Step 1
Identify the coefficient of the $y$ term: $E = -8$.
Step 2
Apply the center formula for the y-coordinate: $k = -\frac{E}{2}$.
Step 3
Substitute the value: $k = -\frac{-8}{2}$.
Step 4
Calculate final coordinate: $k = 4$.
⚡ Speed / Desmos Tactic:
Step 1
Graph $x^2 + y^2 + 10x - 8y + 5 = 0$ in Desmos.
Step 2
Click the center point on the graph to reveal coordinates $(-5, 4)$.
Step 3
Extract the y-coordinate: 4.
Concept 4
Concept 4: Circle-Line Intersections & Tangency
Determining points of intersection between a linear equation and a circle equation by substitution or graphical inspection.
Substitute the linear expression for $y$ (or $x$) into the circle equation.
Expand and simplify into a quadratic equation in one variable: $ax^2 + bx + c = 0$.
Use the discriminant ($b^2 - 4ac$) to determine the number of intersection points: $>0$ (two points), $=0$ (tangent, one point), $<0$ (no intersection).
📘 Traditional Algebraic Method
Isolate $y$ in the linear equation, substitute into the circle equation, expand, solve the resulting quadratic equation using factoring or the quadratic formula.
⚡ SAT Speed Trick & Desmos Hack
Enter both the circle equation and the line equation simultaneously into Desmos and click the intersection points to read coordinates instantly.
💡 Worked SAT Archetype Example
Problem: How many times does the line $y = x$ intersect the circle $x^2 + y^2 = 18$?
📘 Step-by-Step Textbook Solution:
Step 1
Substitute $y = x$ into the circle equation: $x^2 + x^2 = 18$.
Step 2
Combine like terms: $2x^2 = 18$.
Step 3
Divide by 2: $x^2 = 9$.
Step 4
Solve for $x$: $x = 3$ and $x = -3$, yielding 2 distinct intersection points.
⚡ Speed / Desmos Tactic:
Step 1
Type $x^2 + y^2 = 18$ into Desmos line 1.
Step 2
Type $y = x$ into Desmos line 2.
Step 3
Click the two visible grey intersection points on the coordinate plane to confirm count = 2.
Practice Questions (18)
Question 1Identifying Center and Radius from Standard Form
Easy
The equation of a circle in the xy-plane is given by $(x - 3)^2 + (y + 5)^2 = 49$. What is the radius of this circle?
Hint: Recall that the standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.
📘 Step-by-Step Algebraic Solution
Step 1
Identify the given equation as $(x - 3)^2 + (y + 5)^2 = 49$.
Step 2
Compare it to the standard circle equation form: $(x - h)^2 + (y - k)^2 = r^2$.
Step 3
Equate the right side to $r^2$: $r^2 = 49$.
Step 4
Take the positive square root to find the radius: $r = \sqrt{49} = 7$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Type the equation directly into Desmos.
Step 2
Look at the right-hand side constant, which represents $r^2$.
Step 3
Take the square root instantly to get $7$.
Question 2Identifying Center and Radius from Standard Form
Easy
Which of the following represents the coordinates of the center of the circle defined by $(x + 2)^2 + (y - 4)^2 = 25$?
Hint: The coordinates of the center are $(h, k)$, found by taking the opposites of the numbers inside the parentheses with $x$ and $y$.
📘 Step-by-Step Algebraic Solution
Step 1
Review the standard equation of a circle: $(x - h)^2 + (y - k)^2 = r^2$.
Step 2
Match the $x$-term: $(x - h)^2 = (x + 2)^2 \implies -h = 2 \implies h = -2$.
Step 3
Match the $y$-term: $(y - k)^2 = (y - 4)^2 \implies -k = -4 \implies k = 4$.
Step 4
Combine the coordinates to get the center: $(-2, 4)$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Graph the circle in Desmos.
Step 2
Click on the center point visually highlighted on the graph.
Step 3
Read off the coordinates $(-2, 4)$ directly.
Question 3Identifying Center and Radius from Standard Form
Medium
A circle in the xy-plane has the equation $3(x - 1)^2 + 3(y + 2)^2 = 75$. What is the radius of the circle?
Hint: Divide the entire equation by the coefficient of the squared binomials to convert it into standard form first.
📘 Step-by-Step Algebraic Solution
Step 1
Start with the given equation: $3(x - 1)^2 + 3(y + 2)^2 = 75$.
Step 2
Divide all terms by $3$ to isolate the squared binomials: $(x - 1)^2 + (y + 2)^2 = 25$.
Step 3
Recognize that $25$ represents $r^2$ in standard form: $r^2 = 25$.
Step 4
Solve for the radius: $r = \sqrt{25} = 5$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Enter $3(x - 1)^2 + 3(y + 2)^2 = 75$ into Desmos.
Step 2
Observe the plotted circle and measure its radius from center to edge.
Step 3
Conclude that the radius is $5$.
Question 4Identifying Center and Radius from Standard Form
Medium
The equation of a circle is $(x - 4)^2 + (y - 3)^2 = r^2$. If the point $(7, 7)$ lies on the circle, what is the value of $r$?
Hint: Substitute the coordinates of the given point $(x, y) = (7, 7)$ into the circle's equation to solve for $r^2$.
📘 Step-by-Step Algebraic Solution
Step 1
Substitute $x = 7$ and $y = 7$ into the equation: $(7 - 4)^2 + (7 - 3)^2 = r^2$.
Step 2
Simplify inside the parentheses: $(3)^2 + (4)^2 = r^2$.
Graph $x^2 + y^2 = 8$ and $y = x + c$ with a slider for $c$ in Desmos.
Step 2
Adjust $c$ until the line just touches the circle at a single point.
Step 3
Read the value $c = 4$.
Question 14Line-Circle Intersections and Tangency
Medium
A circle has the equation $x^2 + y^2 = 25$. A line passing through $(0, 5)$ is tangent to this circle. Which of the following is the slope of this line?
Hint: The point $(0, 5)$ is on the top of the circle $x^2 + y^2 = 25$. Consider the geometric relationship between radius and tangent lines.
📘 Step-by-Step Algebraic Solution
Step 1
Note that the point $(0, 5)$ lies on the circle $x^2 + y^2 = 25$ because $0^2 + 5^2 = 25$.
Step 2
A radius drawn from the origin $(0, 0)$ to the point of tangency $(0, 5)$ lies along the y-axis, which is vertical.
Step 3
A tangent line at any point on a circle is perpendicular to the radius drawn to that point.
Step 4
Since the radius is vertical, the tangent line must be horizontal, meaning its equation is $y = 5$ and its slope is $0$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Plot $x^2 + y^2 = 25$ and the point $(0, 5)$ in Desmos.
Step 2
Visualize the horizontal line $y = 5$ passing through $(0, 5)$ and touching the top of the circle.
Step 3
Conclude the slope is $0$.
Question 15Line-Circle Intersections and Tangency
Hard
The line $y = 2x + b$ is tangent to the circle $(x - 1)^2 + y^2 = 5$. What is the positive value of $b$?
Hint: Substitute $y = 2x + b$ into the circle equation, expand, set the discriminant to zero, and solve for $b$.
Since this matches the right side, point $(4, 2)$ is on the circle.
⚡ Desmos Shortcut / Speed Hack
Step 1
Enter the circle equation into Desmos.
Step 2
Plot the given options as points: $(1,-2), (4,2), (5,1), (-4,2)$.
Step 3
Visually verify which point lies directly on the circle boundary.
Question 18Circle Geometry Properties and Chords
Medium
A circle in the xy-plane has a diameter with endpoints at $(-3, 2)$ and $(5, 6)$. What is the equation of this circle?
Hint: Use the midpoint formula to find the center $(h, k)$, and use the distance formula between the center and one endpoint to find $r^2$.
📘 Step-by-Step Algebraic Solution
Step 1
Find the center $(h, k)$ using the midpoint formula: $h = \frac{-3 + 5}{2} = 1$, $k = \frac{2 + 6}{2} = 4$. Center is $(1, 4)$.
Step 2
Find the radius squared ($r^2$) using the distance formula between $(1, 4)$ and $(5, 6)$: $r^2 = (5 - 1)^2 + (
⚡ Desmos Shortcut / Speed Hack
Official SAT Exam Blueprint & Weightage
Metric
Exam Pattern & Weightage
Question Frequency
1-3 questions per test module
Module 1 Appearance
Medium Frequency (Foundational tests)
Module 2 Appearance
High Frequency (Hard module score differentiator)
Pattern Analysis
College Board frequently tests circle equations by presenting them in non-standard general form, requiring students to either complete the square or leverage graphing calculator tools like Desmos for rapid center and radius identification.
🏛️
Official SAT PYQ Drill Bank (2023–2026)
We are compiling real verified College Board exam patterns with step-by-step Desmos shortcuts for Circle Equations.
Updated for 2026 Testing Season
Revision, Traps & Desmos Syntax
Standard Form of a Circle
$(x - h)^2 + (y - k)^2 = r^2$
Center is at $(h, k)$ and radius is $r$. Watch out for sign flips inside the parentheses.
General Form of a Circle
$x^2 + y^2 + Dx + Ey + F = 0$
Expanded form requiring completing the square to find center coordinates $(-\frac{D}{2}, -\frac{E}{2})$.
🚨 Top SAT Traps & Misconceptions
⚠️ SAT Trap: Radius vs Radius Squared Trap
The right side of the standard circle equation is $r^2$, not $r$. Students frequently forget to take the square root and select the un-squared constant as the radius.
⚠️ SAT Trap: Center Sign Reversal Trap
Given $(x - 3)^2$, the x-coordinate of the center is $+3$, not $-3$. Always reverse the sign of numbers inside binomials when extracting center coordinates.
⚡ Essential Desmos Cheatsheet
🎯 Instant Center and Radius Readout
(x-h)^2 + (y-k)^2 = r^2
Type any valid circle equation into Desmos and click the center point or intercepts to bypass manual algebra entirely.
3-Level Mock Test (30 Questions)
🟢 Level 1: Foundation
10 Qs · Sub-600 Score
🟡 Level 2: Target 700+
10 Qs · 600–740 Score
🔴 Level 3: 800-Mastery
10 Qs · 750–800 Score
Question 1Level 1: Foundation
What is the radius of the circle given by the equation $(x - 3)^2 + (y + 4)^2 = 49$?
Explanation:
Step 1
Recall the standard form of a circle's equation: $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.
Step 2
Compare the given equation $(x - 3)^2 + (y + 4)^2 = 49$ to the standard form to identify that $r^2 = 49$.
Step 3
Solve for the radius $r$ by taking the square root: $r = \sqrt{49} = 7$.
Question 2Level 1: Foundation
What are the coordinates of the center of the circle represented by the equation $(x + 5)^2 + (y - 2)^2 = 36$?
Explanation:
Step 1
Use the standard equation of a circle: $(x - h)^2 + (y - k)^2 = r^2$, where the center is $(h, k)$.
Step 2
Match $(x + 5)^2$ with $(x - h)^2$ to find $h = -5$, and $(y - 2)^2$ with $(y - k)^2$ to find $k = 2$.
Step 3
Conclude that the center of the circle is $(-5, 2)$.
Question 3Level 1: Foundation
Which of the following equations represents a circle with center $(0, 0)$ and radius $9$?
Explanation:
Step 1
Substitute the center $(h, k) = (0, 0)$ and radius $r = 9$ into the standard circle equation.
Step 2
Write out the equation: $(x - 0)^2 + (y - 0)^2 = 9^2$.
Step 3
Simplify the expression to get $x^2 + y^2 = 81$.
Question 4Level 1: Foundation
A circle has the equation $(x - 2)^2 + y^2 = 16$. What is the diameter of this circle?
Explanation:
Step 1
Identify that $r^2 = 16$ from the standard form equation.
Step 2
Find the radius by taking the square root: $r = \sqrt{16} = 4$.
Step 3
Calculate the diameter by doubling the radius: $\text{Diameter} = 2r = 2(4) = 8$.
Question 5Level 1: Foundation
Which of the following points lies on the circle given by the equation $x^2 + y^2 = 25$?
Explanation:
Step 1
Test the given point $(3, 4)$ by substituting $x = 3$ and $y = 4$ into the equation $x^2 + y^2 = 25$.
Step 2
Evaluate the left side: $3^2 + 4^2 = 9 + 16 = 25$.
Step 3
Since the left side equals the right side ($25 = 25$), the point lies on the circle.
Question 6Level 1: Foundation
What is the radius of the circle defined by the equation $(x + 1)^2 + (y - 3)^2 = 10$?
Explanation:
Step 1
Identify $r^2 = 10$ from the standard circle equation.
Step 2
Take the square root of both sides to find $r$.
Step 3
Conclude that $r = \sqrt{10}$.
Question 7Level 1: Foundation
A circle has its center at $(4, -1)$ and a radius of $5$. Which of the following is its equation?
Explanation:
Step 1
Substitute the center $(h, k) = (4, -1)$ and radius $r = 5$ into $(x - h)^2 + (y - k)^2 = r^2$.
Write the equation with center $(3, -2)$ and $r^2 = 25$: $(x - 3)^2 + (y + 2)^2 = 25$.
Question 6Level 2: Target 700+
Which of the following lines is tangent to the circle $x^2 + y^2 = 16$?
Explanation:
Step 1
Identify that the circle $x^2 + y^2 = 16$ is centered at the origin $(0, 0)$ with a radius of $r = 4$.
Step 2
Recall that a line is tangent to a circle if the perpendicular distance from the center to the line equals the radius.
Step 3
The vertical line $x = 4$ is at a distance of $4$ units from the origin, exactly matching the radius, so it is tangent.
Question 7Level 2: Target 700+
If the equation of a circle is $2x^2 + 2y^2 - 8x + 12y - 6 = 0$, what is the radius of the circle?
Explanation:
Step 1
Divide the entire equation by $2$ to make the leading coefficients of $x^2$ and $y^2$ equal to $1$: $x^2 + y^2 - 4x + 6y - 3 = 0$.
Step 2
Complete the square: $(x - 2)^2 - 4 + (y + 3)^2 - 9 - 3 = 0$, which simplifies to $(x - 2)^2 + (y + 3)^2 = 25$.
Step 3
Determine the radius by taking the square root of $r^2 = 25$, yielding $r = 5$.
Question 8Level 2: Target 700+
A circle passes through the origin $(0, 0)$ and has its center at $(6, 8)$. What is the standard equation of this circle?
Explanation:
Step 1
The center is given as $(6, 8)$, so the equation starts as $(x - 6)^2 + (y - 8)^2 = r^2$.
Step 2
Find the radius squared by finding the distance from the center $(6, 8)$ to the origin $(0, 0)$: $r^2 = (6 - 0)^2 + (8 - 0)^2 = 36 + 64 = 100$.
Step 3
Substitute $r^2 = 100$ into the equation to get $(x - 6)^2 + (y - 8)^2 = 100$.
Question 9Level 2: Target 700+
For what value of $c$ does the equation $x^2 + y^2 - 8x + 10y + c = 0$ represent a circle with a radius of $7$?
Explanation:
Step 1
Complete the square for the equation: $(x^2 - 8x + 16) + (y^2 + 10y + 25) = -c + 16 + 25$.
Step 2
Rewrite in standard form: $(x - 4)^2 + (y + 5)^2 = 41 - c$.
Step 3
Set $41 - c = r^2 = 7^2 = 49$, and solve for $c$: $41 - c = 49 \implies c = -8$.
Question 10Level 2: Target 700+
Find the area of the circle given by the equation $x^2 + y^2 + 2x - 4y - 11 = 0$.
Explanation:
Step 1
Complete the square for $x^2 + 2x$ and $y^2 - 4y$: $(x + 1)^2 - 1 + (y - 2)^2 - 4 - 11 = 0$.
Step 2
Simplify to find the radius squared: $(x + 1)^2 + (y - 2)^2 = 16$, which means $r^2 = 16$ and $r = 4$.
Step 3
Calculate the area of the circle using $A = \pi r^2$: $A = \pi (4^2) = 16\pi$.
Question 1Level 3: 800 Mastery
A circle is tangent to both the x-axis and the y-axis, and its center lies in the first quadrant. If the circle passes through the point $(1, 2)$, what is the sum of all possible values of its radius $r$?
Explanation:
Step 1
Since the circle is tangent to both axes in the first quadrant, its center is $(r, r)$ and its equation is $(x - r)^2 + (y - r)^2 = r^2$.
Step 2
Substitute the given point $(1, 2)$ into the equation: $(1 - r)^2 + (2 - r)^2 = r^2$.
For the line to be tangent, the quadratic must have exactly one solution, meaning the discriminant must equal zero: $\Delta = (4b)^2 - 4(5)(b^2 - 5) = 0$.
Step 4
Solve for $b$: $16b^2 - 20b^2 + 100 = 0 \implies -4b^2 = -100 \implies b^2 = 25$, so the positive value is $b = 5$.
Question 3Level 3: 800 Mastery
A circle passes through the points $(1, 7)$, $(7, 5)$, and $(1, -1)$. What is the y-coordinate of the center of this circle?
Explanation:
Step 1
Notice that two points share the same x-coordinate: $(1, 7)$ and $(1, -1)$. The perpendicular bisector of the chord connecting these points is horizontal, passing through the midpoint's y-coordinate: $y = \frac{7 + (-1)}{2} = 3$.
Step 2
Since the center must lie on the perpendicular bisector of any chord, the y-coordinate of the center is immediately $3$.
Step 3
Verification confirms that the center is $(4, 3)$, making the correct y-coordinate $3$.
Question 4Level 3: 800 Mastery
What is the maximum distance from the point $(10, 7)$ to a point on the circle given by $x^2 + y^2 - 4x - 2y = 20$?
Explanation:
Step 1
Complete the square for the circle equation: $(x - 2)^2 - 4 + (y - 1)^2 - 1 = 20 \implies (x - 2)^2 + (y - 1)^2 = 25$.
Step 2
Identify the center $(2, 1)$ and radius $r = 5$.
Step 3
Calculate the distance $d$ from the given point $(10, 7)$ to the center $(2, 1)$: $d = \sqrt{(10 - 2)^2 + (7 - 1)^2} = \sqrt{8^2 + 6^2} = 10$.
Step 4
The maximum distance from a point to a circle is the distance to the center plus the radius: $\text{Max Distance} = d + r = 10 + 5 = 15$.
Question 5Level 3: 800 Mastery
Two circles are given by the equations $(x - 1)^2 + (y - 2)^2 = 4$ and $(x - 7)^2 + (y - 10)^2 = 9$. What is the shortest distance between a point on the first circle and a point on the second circle?
Explanation:
Step 1
Identify the centers and radii of both circles: Circle 1 has center $C_1(1, 2)$ and radius $r_1 = 2$; Circle 2 has center $C_2(7, 10)$ and radius $r_2 = 3$.
Step 2
Calculate the distance $D$ between the two centers: $D = \sqrt{(7 - 1)^2 + (10 - 2)^2} = \sqrt{6^2 + 8^2} = 10$.
Step 3
Find the shortest distance between the circles by subtracting both radii from the distance between centers: $\text{Shortest Distance} = D - r_1 - r_2 = 10 - 2 - 3 = 5$.
Step 4
Wait, let's re-verify: $D = 10, r_1 = 2, r_2 = 3 \implies 10 - 5 = 5$. Let me check options. Option C is 5. Let me re-read carefully. Ah, wait. Let's check options again. Option A is 2, B is 3, C is 5, D is 10. Let's re-calculate distance: $(7-1)^2 + (10-2)^2 = 36 + 64 = 100$, so $D = 10$. $r_1 = 2, r_2 = 3$. Shortest distance is $10 - 2 - 3 = 5$. Wait, why did I put correct_index as 0? Let's fix correct_index to point to 5, which is option C (index 2).
Question 6Level 3: 800 Mastery
A circle passes through the points $(0, 0)$, $(4, 0)$, and $(0, 6)$. What is the equation of this circle?
Explanation:
Step 1
Use the general form of a circle equation: $x^2 + y^2 + Dx + Ey + F = 0$.
Step 2
Substitute $(0, 0)$ to find $F = 0$. Substitute $(4, 0)$ to find $16 + 4D = 0 \implies D = -4$. Substitute $(0, 6)$ to find $36 + 6E = 0 \implies E = -6$.
Step 3
The equation is $x^2 + y^2 - 4x - 6y = 0$, which completes the square to $(x - 2)^2 + (y - 3)^2 = 13$.
Step 4
Since both forms represent the exact same circle, the correct choice is D.
Question 7Level 3: 800 Mastery
Find the area of the region bounded by the circle $(x - 3)^2 + (y + 4)^2 = 25$ that lies entirely above the x-axis ($y \ge 0$).