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Recover a circle and find every intersection

Use equal-distance conditions to recover a circle, then solve and check complete line–circle intersection sets.

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Enlarged diagram

A circle on a coordinate grid is fixed by its centre and radius. Given three points on it, you can rebuild both; given a line, you can find exactly where it meets the circle. This lesson uses perpendicular bisectors and substitution to do each, and the discriminant to count the meeting points.

Recover the centre from equal distances

A diameter gives its centre immediately: average the two endpoint coordinates. An arbitrary chord does not. When three non-collinear points determine a circle, its centre lies on the perpendicular bisector of every chord.

Worked example 1

Rebuild a circle from two chord bisectors

Find the circle through A(6,2), B(1,7) and C(-4,2). Give its centre, radius and both equation forms.

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  1. Use a horizontal chord

    Method

    Find the midpoint of the chord and its perpendicular bisector.

    Reason

    The centre has equal distances to both chord endpoints, so it lies on their perpendicular bisector.

    Working

    Answer

    Chord AC is horizontal, with midpoint (1,2). Its perpendicular bisector is x = 1.

  2. Use an independent chord

    Method

    Find the perpendicular bisector of the sloping chord and intersect the two bisectors.

    Reason

    One bisector leaves a whole line of possible centres. A second independent condition selects one.

    Working

    Answer

    For AB, the midpoint is (7/2,9/2) and the gradient is -1. Its perpendicular bisector has gradient 1:

    y-9/2 = x-7/2, y = x + 1.

    Together with x = 1, this gives centre O(1,2).

  3. Recover the radius and verify every point

    Method

    Calculate the squared distances from the proposed centre.

    Reason

    All three distances must agree; checking only the chord endpoints would not establish the third condition.

    Working

    Answer

    OA² = (6-1)² + (2-2)² = 25,; OB² = (1-1)² + (7-2)² = 25,; OC² = (-4-1)² + (2-2)² = 25.

  4. State the reconstructed circle

    Working

    Answer

    The centre is (1,2) and radius is 5. The two equivalent equations are

    (x-1)² + (y-2)² = 25,; x² + y²-2x-4y-20 = 0.

If a centre and one circle point are given, their squared distance supplies r². For a stated diameter, r² is one quarter of its squared length. Three distinct collinear points cannot lie on one positive-radius circle.

Check your understanding 1

Use two independent centre conditions

A circle passes through A(-2,1), B(4,1) and C(1,4). Find its centre, radius and equation. Show why your centre fits all three points, rather than treating AB as a diameter without justification.

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AB has midpoint (1,1), so its perpendicular bisector is x = 1. AC has midpoint (-1/2,5/2) and gradient 1, so its perpendicular bisector is y = -x + 2. They meet at (1,1). All three given points are 3 units from this centre, giving (x-1)² + (y-1)² = 9. Here AB happens to be a diameter, but the second condition establishes that rather than assuming it.

An intersection must satisfy both equations

A point belongs to the line and the circle simultaneously. Substitution reduces those two conditions to one quadratic. Each real root supplies one coordinate; use the line to recover its partner.

Practise: G2 Additional Mathematics Line conditions, coordinate regions and single-circle geometry practice · G2 Additional MathematicsPractise: G3 Additional Mathematics Coordinate geometry and straight-line transformations practice · G3 Additional Mathematics

Worked example 2

Find the complete intersection set

Find every intersection of x² + y² = 25 and y = 3.

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  1. Enforce both conditions

    Method

    Replace y by 3 in the circle equation.

    Reason

    Every common point must have the line’s y-coordinate and the circle’s squared distance.

    Working

    Answer

    x² + 3² = 25,; x² = 16,; x = 4 or -4.

    Taking only the positive square root loses one point. Pair each x-value with y = 3: (4,3) and (-4,3).

  2. Check the original equations

    Method

    Test each complete pair in the line and the circle.

    Reason

    A root is useful only when the recovered pair satisfies both original conditions.

    Working

    Answer

    Both have y = 3. Also, 4² + 3² = 25 and (-4)² + 3² = 25. The complete set is {(4,3),(-4,3)}; its order does not matter.

Two, one or no real intersections

Compare three lines on one circle

One circle of radius five about the origin, with horizontal lines at heights three, five and six. The height-five line touches the top; the height-six line is above it.

One circle of radius five about the origin, with horizontal lines at heights three, five and six. The height-five line touches the top; the height-six line is above it.One circle of radius five about the origin, with horizontal lines at heights three, five and six. The height-five line touches the top; the height-six line is above it.
For x² + y² = 25, the line y = 3 cuts the circle twice, y = 5 meets it once and y = 6 has no real intersection. Both axes use the same coordinate scale.
Open full-size graphOpen full-size graph
What to noticeThe sampled circle guides the picture. Substitution into the original equations establishes exact roots and complete paired points. Lines continue beyond the plotting window.
Figure data

This table is included in print while it is open.

Figure data: Compare three lines on one circle
Seriesx (coordinate units)y (coordinate units)
Circle5.00000.0000
Circle4.99310.2617
Circle4.97260.5226
Circle4.93840.7822
Circle4.89071.0396
Circle4.82961.2941
Circle4.75531.5451
Circle4.66791.7918
Circle4.56772.0337
Circle4.45502.2700
Circle4.33012.5000
Circle4.19342.7232
Circle4.04512.9389
Circle3.88573.1466
Circle3.71573.3457
Circle3.53553.5355
Circle3.34573.7157
Circle3.14663.8857
Circle3.00004.0000
Circle2.93894.0451
Circle2.72324.1934
Circle2.50004.3301
Circle2.27004.4550
Circle2.03374.5677
Circle1.79184.6679
Circle1.54514.7553
Circle1.29414.8296
Circle1.03964.8907
Circle0.78224.9384
Circle0.52264.9726
Circle0.26174.9931
Circle0.00005.0000
Circle-0.26174.9931
Circle-0.52264.9726
Circle-0.78224.9384
Circle-1.03964.8907
Circle-1.29414.8296
Circle-1.54514.7553
Circle-1.79184.6679
Circle-2.03374.5677
Circle-2.27004.4550
Circle-2.50004.3301
Circle-2.72324.1934
Circle-2.93894.0451
Circle-3.14663.8857
Circle-3.34573.7157
Circle-3.53553.5355
Circle-3.71573.3457
Circle-3.88573.1466
Circle-4.04512.9389
Circle-4.19342.7232
Circle-4.33012.5000
Circle-4.45502.2700
Circle-4.56772.0337
Circle-4.66791.7918
Circle-4.75531.5451
Circle-4.82961.2941
Circle-4.89071.0396
Circle-4.93840.7822
Circle-4.97260.5226
Circle-4.99310.2617
Circle-5.00000.0000
Circle-4.9931-0.2617
Circle-4.9726-0.5226
Circle-4.9384-0.7822
Circle-4.8907-1.0396
Circle-4.8296-1.2941
Circle-4.7553-1.5451
Circle-4.6679-1.7918
Circle-4.5677-2.0337
Circle-4.4550-2.2700
Circle-4.3301-2.5000
Circle-4.1934-2.7232
Circle-4.0451-2.9389
Circle-3.8857-3.1466
Circle-3.7157-3.3457
Circle-3.5355-3.5355
Circle-3.3457-3.7157
Circle-3.1466-3.8857
Circle-2.9389-4.0451
Circle-2.7232-4.1934
Circle-2.5000-4.3301
Circle-2.2700-4.4550
Circle-2.0337-4.5677
Circle-1.7918-4.6679
Circle-1.5451-4.7553
Circle-1.2941-4.8296
Circle-1.0396-4.8907
Circle-0.7822-4.9384
Circle-0.5226-4.9726
Circle-0.2617-4.9931
Circle0.0000-5.0000
Circle0.2617-4.9931
Circle0.5226-4.9726
Circle0.7822-4.9384
Circle1.0396-4.8907
Circle1.2941-4.8296
Circle1.5451-4.7553
Circle1.7918-4.6679
Circle2.0337-4.5677
Circle2.2700-4.4550
Circle2.5000-4.3301
Circle2.7232-4.1934
Circle2.9389-4.0451
Circle3.1466-3.8857
Circle3.3457-3.7157
Circle3.5355-3.5355
Circle3.7157-3.3457
Circle3.8857-3.1466
Circle4.0451-2.9389
Circle4.1934-2.7232
Circle4.3301-2.5000
Circle4.4550-2.2700
Circle4.5677-2.0337
Circle4.6679-1.7918
Circle4.7553-1.5451
Circle4.8296-1.2941
Circle4.8907-1.0396
Circle4.9384-0.7822
Circle4.9726-0.5226
Circle4.9931-0.2617
Circle5.00000.0000
y = 3-6.00003.0000
y = 36.00003.0000
y = 5-6.00005.0000
y = 56.00005.0000
y = 6-6.00006.0000
y = 66.00006.0000

For the same circle, y = 5 gives x² = 0: the single point (0,5). A repeated quadratic root is one point, not two distinct intersections. The radius there is vertical and the tangent horizontal.

For y = 6, substitution gives x² = -11, so there is no real intersection. An empty point set is a complete answer; it is different from leaving the response blank. In a general substituted quadratic, a positive, zero or negative discriminant distinguishes two, one or no real roots. Always recover and check the corresponding coordinate pairs.

Keep the pairs together on an oblique line

Practise: G2 Additional Mathematics Line conditions, coordinate regions and single-circle geometry practice · G2 Additional MathematicsPractise: G3 Additional Mathematics Coordinate geometry and straight-line transformations practice · G3 Additional Mathematics

Try it yourself 1

Solve with a sloping line

Minimal support

Find every real intersection of x² + y² = 25 and y = x + 1. Show the equation you solve, give complete coordinate pairs and check them in both original equations.

Write your answer, then compare it with the solution

Hint 1

Let one original equation eliminate a coordinate

Use the line to replace y in the circle. After solving, recover y separately for each x-value; do not cross-pair the coordinates.

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  1. Substitute and solve

    Method

    Use y = x + 1 in the circle.

    Reason

    The roots must represent points common to both equations.

    Working

    Answer

    x² + (x + 1)² = 25,; 2x² + 2x-24 = 0,; (x + 4)(x-3) = 0.

    Thus x = -4 or x = 3. The line gives y = -3 or y = 4 respectively, so the set is {(-4,-3),(3,4)}.

  2. Verify the paired answers

    Method

    Test each pair rather than only the separate coordinate lists.

    Reason

    Cross-pairing can satisfy neither original equation.

    Working

    Answer

    For (-4,-3), -3 = -4 + 1 and 16 + 9 = 25. For (3,4), 4 = 3 + 1 and 9 + 16 = 25. Both points satisfy both equations. Reversing their order changes nothing. The pairs (-4,4) and (3,-3) fail the line condition.

Common mistakes

  • Using the midpoint of any chord as the centre. Only a diameter’s midpoint is the centre.
  • Splitting coordinates from their pairs. Each x found gives its own y; report complete points.
  • Counting a repeated root as two points. A discriminant of zero means one point, where the line is a tangent.
  • Dropping the negative root. x² = 9 gives x = 3 or x = -3: two intersection points, each with its own y.

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