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Circles

A circle is one sentence of algebra, every point at distance r from a fixed centre, and Pythagoras turns that sentence into an equation. Completing the square reads centre and radius out of a scrambled form, and three circle theorems from GCSE come back with coordinates and jobs to do.

Year 12-13EDEXCEL 9MA0 3.2

Builds on Quadratic functions and Straight lines.

IN THIS TOPIC

  • Convert between centre-radius and expanded forms of a circle's equation by completing the square.
  • Use the tangent, chord and semicircle properties as coordinate calculations.
  • Find tangent equations at a point, and the circle through three given points.

WHAT YOU PROBABLY THINK

A line that meets a circle meets it twice.

From Pythagoras to the equation

A circle is the set of points at a fixed distance r from a centre (a, b), and the distance between points is Pythagoras. Squaring the distance condition gives the equation directly,

(x − a)2 + (y − b)2 = r2

with the centre and radius on display. Expanding scrambles them into the general form x2 + y2 + 2fx + 2gy + c = 0, and the route back is completing the square in x and y separately, the same move the quadratic functions lesson used to find a vertex.

WORKED EXAMPLE

Unscrambling a general form

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

Complete the square in each variable: (x − 3)2 − 9 + (y + 2)2 − 4 − 12 = 0.

Gather the constants: (x − 3)2 + (y + 2)2 = 25.

Centre (3, −2), radius 5. The signs flip on the way out of the brackets, and the radius is √25, not 25; both slips are marked for.

Tangents and chords

Two GCSE circle facts become calculating tools here. The radius to a point of contact is perpendicular to the tangent there, so a tangent's gradient is the negative reciprocal of the radius gradient, and last lesson finishes the job.

WORKED EXAMPLE

A tangent, without calculus

Find the equation of the tangent to (x − 3)2 + (y + 2)2 = 25 at the point (6, 2).

Confirm the point is on the circle: 32 + 42 = 25. It is.

The radius from (3, −2) to (6, 2) has gradient 4/3, so the tangent's gradient is −3/4.

Through (6, 2): y − 2 = −3/4(x − 6), which clears to 3x + 4y − 26 = 0.

No differentiation appeared, and none is needed; the geometry carries the whole question.

The circle with centre 3 comma minus 2 and radius 5, with the tangent at the point 6 comma 2: the radius to the point has gradient four thirds, the tangent gradient minus three quarters, and they meet at a right anglecentre (3, −2)(6, 2)3x + 4y = 26radius ⊥ tangent at the point of contact
FIG. 1The tangent at (6, 2), built from the right angle. The radius has gradient 4/3, the tangent −¾, and their product is −1.

The second fact says the perpendicular from the centre bisects a chord, which sets up a right triangle: half the chord, the distance from the centre, and the radius as hypotenuse. And substituting a line into a circle's equation gives a quadratic whose discriminant settles the opening lie, two intersections when it is positive, none when negative, and a single touch, a tangent, when it is zero.

A chord of the circle with centre 3 comma minus 2 and radius 5, running from 6 comma 2 down to 6 comma minus 6: the perpendicular from the centre meets it at its midpoint, and the half chord of 4 sits in a 3 4 5 trianglecentre (3, −2)(6, 2)(6, −6)345the perpendicular from the centre lands on the midpoint
FIG. 2The chord from (6, 2) to (6, −6). The perpendicular from the centre meets it at its midpoint, and the 3-4-5 triangle gives the half-chord.

YOUR TURN

A chord measured by Pythagoras

The line x = 6 cuts the circle (x − 3)2 + (y + 2)2 = 25 in a chord. Find the chord's length, before opening the working.

Show the working

Substituting x = 6 gives (y + 2)2 = 16, so y = 2 or y = −6, and the chord runs from (6, 2) to (6, −6), length 8.

The theorem route agrees without solving anything. The centre is 3 from the line x = 6, the radius is 5, so the half-chord is √(25 − 9) = 4.

Two independent methods reaching one answer is the strongest check available in coordinate work.

The angle in a semicircle

The third revived theorem: an angle in a semicircle is a right angle. If AB is a diameter, every other point of the circle sees AB at 90°, and the converse runs the other way, a right angle at C means AB is a diameter of the circle through A, B and C. That converse is the fast route to many circumcircle questions, because a diameter hands over the centre (its midpoint) and the radius (half its length) at once.

The angle in a semicircle: A at minus 1 comma 2 and B at 7 comma 8 are ends of a diameter, so the angle at C, 7 comma 2, on the circle is a right angle, and the circle through all three has centre 3 comma 5 and radius 5A (−1, 2)B (7, 8)C (7, 2)diameterthe angle in a semicircle
FIG. 3A(−1, 2) and B(7, 8) are ends of a diameter, so the angle at C(7, 2) is a right angle, and the circle through all three has centre (3, 5) and radius 5.

TRY IT UNSEEN

A circumcircle by right angle

Find the equation of the circle through A(−1, 2), B(7, 8) and C(7, 2).

Show the working

Look at C first. CA is horizontal and CB is vertical, so the angle at C is a right angle.

By the converse of the semicircle theorem, AB is a diameter. Centre = midpoint of AB = (3, 5); r2 = ((7 + 1)2 + (8 − 2)2)/4 = 100/4 = 25.

The circle is (x − 3)2 + (y − 5)2 = 25, and substituting C gives 16 + 9 = 25, confirming all three points lie on it.

Without the right angle the fallback is honest work: perpendicular bisectors of two chords, intersecting at the centre.

THE EXAM BIT

  • Complete the square in x and y separately and watch both signs: (x − 3)2 means centre x = +3, and r is the square root of the right-hand side.
  • Tangent questions are gradient questions. Radius gradient, negative reciprocal, point-gradient form, done; calculus is never required and often penalised as a method here.
  • Chord lengths come from the half-chord right triangle: (half chord)2 = r2 − d2, with d the centre-to-line distance.
  • Before hunting a circumcircle centre, test the vertices for a right angle; if one appears, the hypotenuse is a diameter and the question is nearly over.
  • Substituting a line into a circle gives a quadratic: discriminant positive for two meetings, zero for a tangent, negative for a miss. Say which case you are in.

CHECK YOURSELF

Find the centre and radius of the circle x2 + y2 + 8x − 2y + 8 = 0, and state whether the point (−1, 1) lies on it.

Show a hint

Complete the square in each variable, then substitute the point.

Show the answer

(x + 4)2 − 16 + (y − 1)2 − 1 + 8 = 0, so (x + 4)2 + (y − 1)2 = 9.

Centre (−4, 1), radius 3.

At (−1, 1): (3)2 + 0 = 9, so the point lies on the circle, at the right-hand end of a horizontal radius.

A circle is Pythagoras with a fixed centre; completing the square recovers centre and radius from any form.

Radius ⊥ tangent, centre-perpendicular bisects chords, and a right angle on the circle names a diameter.

CHECK YOUR PROGRESS

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  • Convert between centre-radius and expanded forms of a circle's equation by completing the square.
  • Use the tangent, chord and semicircle properties as coordinate calculations.
  • Find tangent equations at a point, and the circle through three given points.

No animated video for this topic yet; these notes stand alone.