HSG.GPE.A.1: Deriving the Equation of a Circle and Completing the Square
In plain English: HSG.GPE.A.1 is the Common Core geometry standard that asks students to derive the equation of a circle, (x - h)² + (y - k)² = r², by applying the Pythagorean Theorem to a point on the circle and its center. Students also complete the square on an expanded equation to find the circle's center and radius. It is usually taught in Geometry, sometimes in Algebra II.
Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation.
Common Core State Standards for Mathematics · Domain: Expressing Geometric Properties with Equations (GPE) · Cluster: Translate between the geometric description and the equation for a conic section Also written as HSG-GPE.A.1 or G-GPE.1 · Official standard
Students build the equation of a circle from a single fact: every point on the circle is the same distance r from the center. Placing the center at (h, k) and a point P(x, y) on the circle, they draw a right triangle whose legs are the horizontal and vertical distances |x - h| and |y - k| and whose hypotenuse is the radius. The Pythagorean Theorem then gives (x - h)² + (y - k)² = r².
In the second half, students run the process backward. Given an expanded equation such as x² + y² + Dx + Ey + F = 0, they group the x-terms and y-terms, complete the square in each, and read the center and radius from the result. They also learn to recognize when the right side is zero or negative, so the equation does not describe a circle at all.
Learning Objectives
By the end of this lesson, students will be able to:
Derive the equation (x - h)² + (y - k)² = r² for a circle with center (h, k) and radius r by applying the Pythagorean Theorem to a right triangle
Write the equation of a circle from its center and radius, or from its center and one point on the circle
Complete the square in both x and y to rewrite an expanded circle equation in center-radius form
Identify the center and radius from center-radius form, including the sign change for h and k
Recognize when an equation of the form x² + y² + Dx + Ey + F = 0 has no real points or only one point
Prior Knowledge Required
Students should already be comfortable with:
The Pythagorean Theorem and its use for distances between points 8.G.B.8
Squaring binomials such as (x - 3)² = x² - 6x + 9
Completing the square in a quadratic expression HSA.SSE.B.3
Plotting points and reading coordinates in all four quadrants
Students work alone on graph paper for three minutes, then compare with a partner.
Warm-Up Prompt
"How far is the point (3, 4) from the origin? How far is (-4, 3)? Find four more points with integer coordinates that are exactly 5 units from the origin. If you plotted every point that is 5 units from the origin, what shape would you get?"
Both distances are 5, because 3² + 4² = 25 and (-4)² + 3² = 25. Collect points on the board: (5, 0), (0, -5), (-3, -4), (4, -3) and others. Ask students how they tested each point, and write their test as x² + y² = 25. Tell students that this is already the equation of a circle, and that today they will build the same kind of test for a circle whose center is anywhere.
Direct Instruction20 minutes
Part 1: Deriving the equation. Show Diagram 1. Mark the center C(h, k) and a general point P(x, y) on the circle. Walk through the derivation:
Draw the right triangle: from C, move horizontally to the point (x, k), then vertically up or down to P. The corner at (x, k) is a right angle.
Name the legs: the horizontal leg has length |x - h| and the vertical leg has length |y - k|. The hypotenuse is CP, which is the radius r.
Apply the Pythagorean Theorem: |x - h|² + |y - k|² = r². Because squaring removes the sign, this is (x - h)² + (y - k)² = r².
Argue both directions: every point on the circle satisfies the equation, and any point that satisfies it is r units from C, so it is on the circle. The equation describes the circle exactly.
Point out that the derivation does not depend on where P is. If P is to the left of or below the center, the differences x - h or y - k are negative, but their squares are still the squared leg lengths. If P is directly beside or above the center, one leg has length 0 and the equation still holds.
Part 2: From an expanded equation back to center and radius. Expand (x - 4)² + (y + 3)² = 36 with the class to get x² + y² - 8x + 6y - 11 = 0, then show with Diagram 2 how completing the square reverses that expansion. Work the examples below.
Center and radius given
Write the equation of the circle with center (-3, 2) and radius 6.
Equation: (x + 3)² + (y - 2)² = 36
Center and one point given
The circle is centered at the origin and passes through (5, -12). The legs are 5 and 12, so r² = 25 + 144.
Equation: x² + y² = 169, radius 13
Completing the square
Find the center and radius of x² + y² - 8x + 6y - 11 = 0. Add 16 and 9 to both sides.
Equation: (x + 2)² + (y - 1)² = -4: no real points
For the last example, ask: can a sum of two squares equal -4? No, so no point (x, y) satisfies the equation. Mention that if the right side had come out 0, the only solution would be the center itself. Only a positive right side gives a circle.
Guided Practice15 minutes
Pairs work four problems on whiteboards. After each problem, one pair explains its reasoning before the class moves on.
Guided practice problems and answers
Problem
Answer
Center (1, -4), radius 3
(x - 1)² + (y + 4)² = 9
Center (-2, -5), passes through (1, -1)
Legs 3 and 4, r = 5: (x + 2)² + (y + 5)² = 25
x² + y² + 10x - 4y + 20 = 0
(x + 5)² + (y - 2)² = 9: center (-5, 2), radius 3
x² + y² - 6y = 16
x² + (y - 3)² = 25: center (0, 3), radius 5
For the second problem, insist that students sketch the right triangle between the center and the given point before computing. Listen for these errors: writing (x - 2) for a center with h = -2, stopping at r² and calling it the radius, adding 25 to the left side but not the right side, and trying to complete the square on a missing x-term in the last problem (there is nothing to complete when the x-coefficient is 0).
Independent Practice15 minutes
Students work alone on five problems and check each answer by substituting one point that should be on the circle.
Write the equation of the circle with center (0, 7) and radius √11. (x² + (y - 7)² = 11)
A diameter has endpoints (-1, 2) and (7, 8). Find the center, the radius and the equation. (Center (3, 5), diameter 10, radius 5, (x - 3)² + (y - 5)² = 25)
Find the center and radius of x² + y² - 2x + 14y + 25 = 0. ((x - 1)² + (y + 7)² = 25: center (1, -7), radius 5)
Find the center and radius of 3x² + 3y² - 18x + 12y - 9 = 0. (Divide by 3, then (x - 3)² + (y + 2)² = 16: center (3, -2), radius 4)
Is the point (4, 1) on the circle (x - 1)² + (y + 3)² = 25? Use the right triangle to explain. (Yes: legs 3 and 4, so the distance is 5)
Closure5-10 minutes
Exit ticket: (1) In two sentences, explain where the two squared terms in (x - h)² + (y - k)² = r² come from. (2) Find the center and radius of x² + y² + 2x - 8y + 8 = 0. (Answer: (x + 1)² + (y - 4)² = 9, center (-1, 4), radius 3.) Collect the tickets and sort them by whether the explanation mentions the right triangle and the legs.
Differentiation Strategies
For Struggling Students
Provide a template with blanks: center ( __ , __ ), horizontal leg x - __ , vertical leg y - __ , radius __ , so students see where h, k and r go
Have students circle the x-terms and box the y-terms before completing the square, and write the number they add above each group
Let students plot the center and one point on graph paper and count the leg lengths before using the formula
For Advanced Students
Show that x² + y² + Dx + Ey + F = 0 is a circle exactly when D² + E² - 4F > 0, and find the center and radius in terms of D, E and F
Find the equation of the circle through the three points (0, 0), (8, 0) and (0, 6), and explain why the center is the midpoint of the segment from (8, 0) to (0, 6)
Explain why an equation such as x² + 4y² - 6x = 7 cannot be a circle, even after completing the square
Assessment Guidance
What to Look For
Look for derivations that name the right triangle, identify the legs as |x - h| and |y - k|, and state the Pythagorean Theorem before writing the equation. A student who only memorized the formula will usually skip the triangle. When students complete the square, check that they add the same numbers to both sides, divide by the common leading coefficient first, and report the radius as the square root of the right side. Ask every student to check one point by substitution: this catches most sign errors in h and k.
02
Classroom Activities
3 Activities
1
Twelve Lattice Points
20 minPairs
Students find every point with integer coordinates on one circle, draw the right triangle for several of them, and generalize the pattern to a point (x, y). The derivation comes from their own triangles rather than from a formula on the board.
Procedure
Each pair plots the center (-1, -2) on graph paper and uses a compass to draw the circle of radius 5
Pairs find all 12 points on the circle with integer coordinates: four at the ends of the horizontal and vertical radii, such as (4, -2) and (-1, 3), and eight where the legs are 3 and 4, such as (2, 2) and (3, 1)
For three of the eight points, pairs draw the right triangle to the center and write the Pythagorean statement, for example 3² + 4² = 5²
Pairs replace the numbers with a general point (x, y): the legs are |x + 1| and |y + 2|, so (x + 1)² + (y + 2)² = 25
Pairs test (3, 2) and (1, 0) with the equation and decide whether each is on, outside or inside the circle ((3, 2) gives 32, outside; (1, 0) gives 8, inside)
Discussion Questions
Why are the leg lengths written with absolute value, and why does the final equation not need it?
What does the left side of the equation measure for a point that is not on the circle?
Why does the equation use + 1 and + 2 when the center is (-1, -2)?
Modification for Distance Learning
Use a free graphing tool with a movable point on the circle. Students drag the point, record the leg lengths shown by the tool in a shared table, and check that the sum of their squares is always 25.
2
Expanded or Center-Radius? Card Match
20 minGroups of 3-4
Groups receive 8 cards: 4 expanded equations and 4 answer cards showing a center-radius equation, a center and a radius. Groups complete the square to match each equation with its answer, then check each match by expanding.
Shuffle the 8 cards and deal them face up. Each student takes one equation card and completes the square on paper
The group finds the answer card that matches each result. A second student checks the match by expanding the center-radius equation back to the expanded form
Groups graph all four circles on one set of axes and label each center
Challenge Variation
Add a fifth equation card with no answer card: x² + y² - 2x + 4y + 7 = 0. Groups must explain why no answer card fits (completing the square gives (x - 1)² + (y + 2)² = -2, which no point satisfies).
3
Coverage Maps: Towers and Sprinklers
20 minPairs
Students use circle equations to describe a coverage boundary on a map grid. In the first task they derive the equation from a center and radius; in the second they complete the square on a given boundary equation to find where the device is and how far it reaches.
Task A: A cell tower
A town map uses a grid with 1 unit = 1 kilometer. A cell tower at (3, -2) reaches every point within 6 km
Pairs sketch the right triangle from the tower to a general point (x, y) on the edge of coverage and write the boundary equation: (x - 3)² + (y + 2)² = 36
Pairs decide whether three homes are inside, on or outside the boundary: (7, 2) gives 32 (inside), (-2, 2) gives 41 (outside), (3, 4) gives 36 (on the edge)
Task B: A garden sprinkler
A garden plan uses a grid in meters. The edge of the watered region is x² + y² - 10x + 4y + 13 = 0
Pairs complete the square to find the sprinkler's location and reach: (x - 5)² + (y + 2)² = 16, so the sprinkler is at (5, -2) and waters up to 4 m away
Pairs decide whether a rose bed at (8, 1) is watered (9 + 9 = 18, which is more than 16, so no) and whether a bench at (7, 0) gets wet (4 + 4 = 8, so yes)
Discussion Questions
Comparing the left side with r² tells you inside or outside. Why does that work?
Which task needed the Pythagorean Theorem directly, and which needed completing the square?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Deriving the Circle Equation from a Right Triangle
The circle with center C(2, 1) and radius 5, drawn to scale on a unit grid. For the point P(5, 5), the horizontal leg is 3, the vertical leg is 4 and the hypotenuse is the radius, so 3² + 4² = 5². Replacing 5 and 5 with a general point (x, y) gives (x - 2)² + (y - 1)² = 25.
Diagram 2: Completing the Square to Find Center and Radius
The expanded equation x² + y² - 8x + 6y - 11 = 0 becomes (x - 4)² + (y + 3)² = 36 after adding 16 and 9 to both sides. The graph, drawn to scale, shows the center (4, -3) and a radius of 6.
04
Homework Assignment
~30 min
HSG.GPE.A.1 Homework: Equations of Circles
Directions: Show all work. In Part 1, sketch the right triangle between the center and a point on the circle before you write an equation. In Part 2, show each number you add to both sides. Check every answer by substituting one point on the circle.
Part 1: Deriving the Equation (Problems 1-3)
Write the equation of the circle with center (5, -1) and radius 7. Then sketch a point (x, y) on the circle, label the legs of the right triangle, and use the Pythagorean Theorem to explain why every point on the circle satisfies your equation.
A circle has center (-4, 3) and passes through the point (2, 11). Use the Pythagorean Theorem to find the radius, then write the equation of the circle.
The endpoints of a diameter of a circle are (-3, 4) and (5, -2). Find the center and the radius, and write the equation of the circle.
Part 2: Completing the Square (Problems 4-6)
Find the center and radius of the circle x² + y² + 14x - 2y + 34 = 0.
Find the center and radius of the circle 4x² + 4y² - 16x - 24y + 27 = 0. Give the radius as a fraction.
A student says that x² + y² - 6x + 10y + 40 = 0 is a circle with center (3, -5) and radius √40. Complete the square, explain the student's error, and decide whether the equation describes a circle at all.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Derivation
Right triangle drawn, legs labeled, Pythagorean Theorem stated and used
Equation correct but triangle or reasoning incomplete
No derivation
Completing the Square
Same numbers added to both sides, correct center-radius form
Correct method with one arithmetic or sign error
Method missing or incorrect
Center and Radius
Center signs and radius (not r²) correct in every problem
One sign or radius error
Several errors
Checking and Reasoning
Point checked in each problem; Problem 6 error explained clearly
Some checks or explanation missing
No checks or explanation
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Answer the questions in order. Your score updates as you go, and Reset quiz clears all answers so you or your students can start again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Which equation describes the circle with center (3, -5) and radius 4?
Answer: A
With h = 3 and k = -5, the legs are x - 3 and y - (-5) = y + 5, and r² = 16. Choice B flips both signs of the center. Choice C uses the radius 4 instead of r² = 16 on the right side. Choice D gets the sign of k wrong.
Question 2 of 20 · Multiple Choice
To derive the equation of a circle with center (h, k), you draw a right triangle from the center to a point P(x, y) on the circle. What are the lengths of the two legs?
Answer: B
The legs are the horizontal and vertical distances between P and the center: |x - h| and |y - k|. The radius is the hypotenuse, not a leg, so choice D is wrong. Choice A measures from the origin instead of from the center, which only works when the center is (0, 0).
Question 3 of 20 · Multiple Choice
A circle has center (-2, 6) and passes through the point (10, 1). What is its radius?
Answer: C
The legs are 10 - (-2) = 12 and 1 - 6 = -5, so r² = 144 + 25 = 169 and r = 13. Choice B adds the legs instead of using the Pythagorean Theorem. Choice D is r², not r. Choice A comes from a sign error: using 10 - 2 = 8 and 1 + 6 = 7.
Question 4 of 20 · Multiple Choice
What is the equation of the circle centered at the origin that passes through (-7, 24)?
Answer: B
The legs are 7 and 24, so r² = 49 + 576 = 625 (r = 25). Choice C writes the radius 25 on the right side without squaring it. Choice D uses the given point as the center. Choice A adds 7 and 24.
Question 5 of 20 · Multiple Choice
Which point lies on the circle (x - 1)² + (y + 2)² = 100?
Answer: D
For (7, 6): (7 - 1)² + (6 + 2)² = 36 + 64 = 100, so it is on the circle. Choice A gives 36 + 16 = 52; it would be on the circle only if the center were (1, 2), a sign error in k. Choice B gives 25 + 25 = 50, and choice C gives 1 + 100 = 101, just outside.
Question 6 of 20 · Multiple Choice
A diameter of a circle has endpoints (2, -3) and (8, 5). What is the equation of the circle?
Answer: A
The center is the midpoint (5, 1). The diameter has legs 6 and 8, so its length is 10 and the radius is 5, giving r² = 25. Choice B squares the diameter instead of the radius. Choice C has the wrong signs for the center. Choice D uses half the differences of the coordinates (3 and 4) as the center instead of the midpoint.
Question 7 of 20 · Multiple Choice
What number must be added to x² - 10x to make a perfect square trinomial?
Answer: C
Take half of -10 and square it: (-5)² = 25, and x² - 10x + 25 = (x - 5)². Choice A halves but does not square. Choice D squares 10 without halving it first.
Question 8 of 20 · Multiple Choice
What are the center and radius of x² + y² + 6x - 14y + 42 = 0?
Answer: C
Move 42 and add 9 and 49 to both sides: (x + 3)² + (y - 7)² = -42 + 9 + 49 = 16. The center is (-3, 7) and the radius is √16 = 4. Choice A reads the signs inside the parentheses as the center. Choice B reports r² as the radius. Choice D treats the constant 42 as r².
Question 9 of 20 · Multiple Choice
What are the center and radius of x² + y² - 4y = 21?
Answer: D
There is no x-term, so x² stays as it is. Add 4 to both sides: x² + (y - 2)² = 25. The center is (0, 2) and the radius is 5. Choice B forgets to add 4 to the right side. Choice A has the wrong sign for k, and choice C puts the shift on the wrong variable.
Question 10 of 20 · Multiple Choice
What are the center and radius of 3x² + 3y² - 12x + 18y - 9 = 0?
Answer: A
Divide by 3: x² + y² - 4x + 6y - 3 = 0. Then (x - 2)² + (y + 3)² = 3 + 4 + 9 = 16, so the center is (2, -3) and the radius is 4. Choice C completes the square without dividing by 3 first. Choice B stops after dividing and uses 3 as r² without adding 4 and 9. Choice D flips the signs of the center.
Question 11 of 20 · Multiple Choice
Which equation has no real points, so it does not describe a circle?
Answer: B
Choice B becomes (x + 4)² + y² = -17 + 16 = -1. A sum of squares cannot be negative, so no point satisfies it. The others are circles: A is (x - 1)² + (y - 2)² = 9, C is x² + (y + 3)² = 9 and D is (x - 5)² + (y + 5)² = 25. Choice C may look odd because the constant is 0, but completing the square still gives a positive right side.
Question 12 of 20 · Multiple Choice
A student completes the square on x² + y² + 8x - 2y = 8. What total must be added to both sides?
Answer: D
Add (8/2)² = 16 for the x-terms and (-2/2)² = 1 for the y-terms, 17 in all. The result is (x + 4)² + (y - 1)² = 25. Choice A forgets the y-terms. Choice B halves 8 and 2 but does not square. Choice C squares 8 and 2 without halving them.
Question 13 of 20 · Multiple Choice
A student writes the circle with center (-4, 0) and radius 3 as (x - 4)² + y² = 9. What is wrong?
Answer: C
The horizontal leg is x - h = x - (-4) = x + 4, so the correct equation is (x + 4)² + y² = 9. The student's equation has center (4, 0). Choice A confuses r with r², and choice D puts the radius into the y-term.
Question 14 of 20 · Multiple Choice
Which equation is x² + y² - 2x + 12y + 21 = 0 written in center-radius form?
Answer: A
Move 21 and add 1 and 36 to both sides: (x - 1)² + (y + 6)² = -21 + 1 + 36 = 16. Choice B adds 21 instead of moving it to the other side. Choice C flips the signs, and choice D does not halve the coefficients.
Question 15 of 20 · Short Answer
A circle has center (6, -2) and passes through the point (1, 10). Use the Pythagorean Theorem to find the radius and write the equation of the circle.
The legs are |1 - 6| = 5 and |10 - (-2)| = 12, so r² = 25 + 144 = 169 and r = 13. The equation is (x - 6)² + (y + 2)² = 169.
Question 16 of 20 · Short Answer
Find the center and radius of x² + y² - 16x + 4y + 19 = 0.
Group and move the constant: (x² - 16x) + (y² + 4y) = -19. Add 64 and 4 to both sides: (x - 8)² + (y + 2)² = 49. Center (8, -2), radius 7.
Question 17 of 20 · Short Answer
Explain, step by step, why a circle with center (h, k) and radius r has the equation (x - h)² + (y - k)² = r². Start with a point P(x, y) on the circle.
Draw a horizontal segment from the center to (x, k) and a vertical segment from there to P. They meet at a right angle, so the center, (x, k) and P form a right triangle. The legs have lengths |x - h| and |y - k|, and the hypotenuse is the radius r because P is on the circle. By the Pythagorean Theorem, |x - h|² + |y - k|² = r², and since |a|² = a² for every number a, this is (x - h)² + (y - k)² = r². Conversely, any point satisfying the equation is r units from the center, so it lies on the circle.
Question 18 of 20 · Short Answer
Find the center and radius of 2x² + 2y² + 20x + 8y + 50 = 0.
Divide by 2: x² + y² + 10x + 4y + 25 = 0. Add 25 and 4 to both sides: (x + 5)² + (y + 2)² = -25 + 25 + 4 = 4. Center (-5, -2), radius 2.
Question 19 of 20 · Short Answer
A park map uses a grid in meters. A circular pond has its center at (12, 9), and its edge passes through a bench at (0, 4). Write the equation of the pond's edge and give the pond's radius.
The legs from the center to the bench are 12 and 5, so r² = 144 + 25 = 169 and the radius is 13 meters. The edge is (x - 12)² + (y - 9)² = 169.
Question 20 of 20 · Short Answer
For what value of c is x² + y² - 6x + 8y + c = 0 a circle with radius 6?
Complete the square: (x - 3)² + (y + 4)² = 9 + 16 - c = 25 - c. For radius 6 we need 25 - c = 36, so c = -11.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.GPE.A.1 mean?
HSG.GPE.A.1 means students can build the equation of a circle from its center and radius and can take a circle equation apart again. They derive (x - h)² + (y - k)² = r² from the Pythagorean Theorem, and they complete the square on an expanded equation such as x² + y² + Dx + Ey + F = 0 to find the center and radius.
Is HSG.GPE.A.1 taught in Geometry or Algebra 2?
It is usually taught in high school Geometry, in a unit on coordinate geometry or circles. Some courses revisit it in Algebra II or Precalculus when they study conic sections. The completing-the-square half relies on algebra skills from Algebra I (HSA.SSE.B.3).
Why is the circle equation the Pythagorean Theorem in disguise?
Because the radius is the hypotenuse of a right triangle whose legs are the horizontal and vertical distances from the center to a point on the circle. Writing a² + b² = c² with a = x - h, b = y - k and c = r gives the circle equation directly. The distance formula is the same idea written as a square root.
Why does the equation have minus signs when the center has positive coordinates?
The equation measures differences from the center, x - h and y - k. For a center of (2, 5), those differences are x - 2 and y - 5. For a center of (-2, -5), they are x - (-2) = x + 2 and y + 5. A quick check is to substitute the center: both squared terms should become 0.
How do you complete the square to find the center and radius of a circle?
Group the x-terms and the y-terms, move the constant to the right side, then add the square of half of each linear coefficient to both sides. Factor each group as a perfect square. For example, x² + y² + 2x - 6y = 6 becomes (x + 1)² + (y - 3)² = 16, so the center is (-1, 3) and the radius is 4.
What if the x² and y² terms have a coefficient other than 1?
Divide every term by that coefficient first. If x² and y² have the same coefficient, dividing gives an ordinary circle equation. If they have different coefficients, as in x² + 4y² = 16, the graph is not a circle; that kind of equation belongs to ellipses (HSG.GPE.A.3).
What happens if the right side comes out zero or negative?
If the right side is negative, no point satisfies the equation, because a sum of two squares is never negative. If it is zero, only the center satisfies it, so the graph is a single point. The equation describes a circle only when the right side is positive.
What are common mistakes on circle equation problems?
Frequent errors include: using the wrong signs for the center, reporting r² as the radius, adding the completing-the-square numbers to only one side, forgetting to divide by a common leading coefficient, and squaring the diameter instead of the radius. Checking one point on the circle by substitution catches most of them.
Are circle equations on the SAT?
Yes. Equations of circles in the xy-plane, including rewriting an expanded equation by completing the square to find the center or radius, are part of the Geometry and Trigonometry domain of the digital SAT. State tests aligned to Common Core Geometry also assess this standard.
What comes after HSG.GPE.A.1?
The same method, writing a distance condition and simplifying it, leads to the parabola in HSG.GPE.A.2 and, in advanced courses, to ellipses and hyperbolas in HSG.GPE.A.3. Circle equations are also used to prove or disprove that a point lies on a circle (HSG.GPE.B.4), and the unit circle x² + y² = 1 is the basis of trigonometric functions (HSF.TF.A.2).
07
Related Standards
6 standards
These standards connect to HSG.GPE.A.1: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.G.B.8Prerequisite
Use the Pythagorean Theorem to find the distance between two points on a grid
