The Equation of a Circle · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

The Equation of a Circle

Mathematics · Geometry · ages 16-17
Name ______________________   Date ____________

  1. What is the centre of the circle (x + 6)² + (y - 3)² = 16?

    • (6, -3)
    • (-6, 3)
    • (6, 3)
    • (-6, -3)

  2. A circle has centre (2, 3) and radius 4, so its equation is (x - 2)² + (y - 3)² = k. What is k?

    Answer: ______________

  3. A circle has centre (0, -3) and radius 6. In the equation (x - 0)² + (y + 3)² = r², what number is r²?

    Answer: ______________

  4. What is the centre of the circle (x - 2)² + (y - 5)² = 9?

    • (2, 5)
    • (-2, -5)
    • (5, 2)
    • (2, -5)

  5. Find the centre and radius of the circle x² + y² - 6x + 4y - 12 = 0.

    • Centre (3, -2), radius 5
    • Centre (-3, 2), radius 5
    • Centre (3, -2), radius 25
    • Centre (6, -4), radius 5

  6. Completing the square gives x² - 8x = (x - p)² - q. What is q?

    Answer: ______________

  7. Mia writes the circle with centre (5, -2) and radius 3 as (x + 5)² + (y - 2)² = 9. Where is the slip?

    • the radius should have been squared before it was written down
    • only the first bracket is wrong, so it is (x - 5)² + (y - 2)² = 9
    • both bracket signs are flipped, so it is (x - 5)² + (y + 2)² = 9
    • a negative coordinate cannot appear in a circle equation

  8. Where does the point (1, 7) sit relative to the circle (x - 1)² + (y - 3)² = 25?

    • on the circle
    • inside the circle
    • outside the circle
    • on the circle, because the point shares the centre's x-coordinate

  9. Sam completes the square on x² + y² - 10x + 6y + 9 = 0 and writes (x - 5)² + (y + 3)² = -9. Where is the slip?

    • the -25 and -9 left by the two brackets never crossed over, so the right-hand side is 25
    • the sign of the 6y term was misread, so the second bracket should be (y - 3)²
    • the +9 at the end of the original should have been dropped
    • completing the square works on x alone, never on y

  10. Find the centre and radius of the circle x² + y² - 2x + 6y - 3 = 0.

    • Centre (1, -3), radius √13
    • Centre (-1, 3), radius √13
    • Centre (1, -3), radius 13
    • Centre (1, -3), radius √3
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Answer key

For grown-ups. Fold this page away before handing over the rest.

The Equation of a Circle W1-mt_YARqmi_ioe-s1

  1. (-6, 3) · The bracket (x + 6) is (x - (-6)), so a = -6. The bracket (y - 3) gives b = 3, so the centre is (-6, 3).
  2. 16 · The right-hand side is the radius squared, so k = 4² = 16.
  3. 36 · r² is just the radius multiplied by itself.
  4. (2, 5) · The centre is (a, b) where the brackets read (x - a) and (y - b). Here they read (x - 2) and (y - 5), so the centre is (2, 5).
  5. Centre (3, -2), radius 5 · Completing the square on the x-terms and y-terms turns the equation into (x - 3)² + (y + 2)² = 25, so the centre is (3, -2) and the radius is 5.
  6. 16 · Half of -8 is -4, so the bracket is (x - 4)². Expanding it gives x² - 8x + 16, which is 16 too big, so q = 16.
  7. both bracket signs are flipped, so it is (x - 5)² + (y + 2)² = 9 · Each bracket carries the opposite sign of its coordinate. A centre of (5, -2) gives (x - 5)² + (y + 2)². Mia did square the radius correctly.
  8. inside the circle · Substituting gives (1 - 1)² + (7 - 3)² = 0 + 16 = 16, which is smaller than 25, so the point is inside.
  9. the -25 and -9 left by the two brackets never crossed over, so the right-hand side is 25 · x² - 10x is (x - 5)² - 25 and y² + 6y is (y + 3)² - 9. With the original +9 that leaves -25 on the left, so the right-hand side is 25, not -9.
  10. Centre (1, -3), radius √13 · Completing the square gives (x - 1)² + (y + 3)² = 13, so the radius is √13, not a whole number.
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