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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 recovers centre and radius from a scrambled form, and three circle theorems from GCSE come back with coordinates and real jobs to do.

Year 12-13EDEXCEL 9MA0 3.2

Builds on Quadratic functions and Straight lines.

IN THIS TOPIC

Move between the centre-radius form and the general form by completing the square.
Find the tangent at a point from the gradient of the radius.
Get a chord length from the half-chord right triangle without solving anything.
Find the circle through three given points.

COMMON MISCONCEPTION

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 two points is Pythagoras. Square the distance condition and the equation appears,

(x − a)^{2} + (y − b)^{2} = r^{2}

NOT IN THE BOOKLET — LEARN IT

with the centre and radius on display. The booklet prints no circle equation of any kind, so this one is yours to know. 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 and not 25. Both of those slips are marked for.

Tangents

The first of the revived GCSE facts is the useful one. The radius to a point of contact is perpendicular to the tangent there, so the tangent's gradient is the negative reciprocal of the radius gradient, and last lesson finishes the job. Two positions have no reciprocal to take and get named instead. A horizontal radius meets a vertical tangent, whose equation is x = the x-coordinate of the point, and a vertical radius meets a horizontal tangent y = the y-coordinate.

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, so 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 wanted. The geometry does all of the work.

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.

Chords, and lines that miss

The second fact says the perpendicular from the centre bisects a chord, which builds a right triangle out of half the chord, the distance from the centre to the line, and the radius as hypotenuse. Substituting a line into a circle's equation gives a quadratic instead, and its discriminant settles the geometry with no picture needed. Positive and there are two intersections. Zero and the line touches once, a tangent. Negative and the line misses the circle entirely.

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.

GUIDED PRACTICE

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. 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 and the radius is 5, so the half-chord is √(25 − 9) = 4.

Two independent methods landing on one answer is the strongest check available anywhere in coordinate work.

The angle in a semicircle

The third theorem says an angle in a semicircle is a right angle. If AB is a diameter then every other point of the circle sees AB at 90°, and the converse runs the other way, so a right angle at C means AB is a diameter of the circle through A, B and C. That converse is the fast route through most circumcircle questions, because a diameter gives the centre, its midpoint, and the radius, half its length, at once.

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.

INDEPENDENT PRACTICE

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), and 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 sit on it.

Without a right angle the fallback is the longer method. Perpendicular bisectors of two of the chords, intersected, give the centre.

ASSESSMENT FOCUS

Complete the square in x and y separately and watch both signs. (x − 3)² means the centre has x = +3, and r is the square root of the right-hand side.
In the general form x² + y² + 2fx + 2gy + c = 0 the centre is (−f, −g) and r² = f² + g² − c. Quoting that is faster, but show the completed square whenever the question says “show that”.
Tangent questions are gradient questions. Radius gradient, negative reciprocal, point-gradient form, done. Calculus is never required here and wastes time when it appears.
Chord lengths come from the half-chord right triangle, (half chord)² = r² − d², with d the distance from the centre to the line.
Before hunting a circumcircle centre, test the three vertices for a right angle. If one turns up, the hypotenuse is a diameter and the question is nearly over.
Substituting a line into a circle gives a quadratic, and the discriminant says which case you are in. Two meetings, one touch, or a miss. Name the case in words as well as computing it.

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) the left side is 3² + 0 = 9, so the point lies on the circle, at the right-hand end of a horizontal radius.

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

Radius meets tangent at a right angle, the perpendicular from the centre bisects a chord, and a right angle on the circle names a diameter.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Or read them with their worked answers on the circles questions page.

CHECK YOUR PROGRESS

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Move between the centre-radius form and the general form by completing the square.
Find the tangent at a point from the gradient of the radius.
Get a chord length from the half-chord right triangle without solving anything.
Find the circle through three given points.

Open the full revision checklist to see every objective in the course in one place.