Loci: Apollonius circles and arcs

|z - a| = k|z - b| with k ≠ 1

Put z = x + iy, square both sides and collect terms. Because k ≠ 1, the x² and y² terms do not cancel, and you get a circle. Divide by the coefficient of x², then complete the square to find the centre and radius.

State clearly that the locus is a circle: the scheme of work quotes examiners noting that many lost a mark by not concluding explicitly that the equation was a circle. (When k = 1 it is the perpendicular bisector.)

arg((z - a)/(z - b)) = θ

arg((z - a)/(z - b)) = arg(z - a) - arg(z - b) is the angle at z between the lines to a and to b. By the circle theorem 'angles in the same segment are equal', z moves on an arc of a circle through a and b. The points a and b themselves are not included.

Finding the arc

• The chord ab subtends angle θ at the circumference, so it subtends 2θ at the centre; the radius is |a - b|/(2sin θ).
• θ = π/2 gives a semicircle on ab as diameter.
• To see which side of ab the arc lies, test one point: the angle is measured anticlockwise from the direction of z - b to the direction of z - a.

Greatest and least values

For a point z on a circle with centre c and radius r, the greatest |z - p| is |c - p| + r and the least is ||c - p| - r|, along the line through p and c.

Worked example

Show that |z - 2| = 2|z + i| is a circle and find its centre and radius.

1. (x - 2)² + y² = 4[x² + (y + 1)²].
2. x² - 4x + 4 + y² = 4x² + 4y² + 8y + 4, so 3x² + 3y² + 4x + 8y = 0.
3. x² + y² + (4/3)x + (8/3)y = 0, so (x + 2/3)² + (y + 4/3)² = 4/9 + 16/9 = 20/9.

Answer: A circle with centre (-2/3, -4/3) and radius 2√5/3

Check yourself

1. Find the radius of the circle |z - 3| = 2|z|.
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   2. (x - 3)² + y² = 4x² + 4y² gives 3x² + 3y² + 6x - 9 = 0, so x² + y² + 2x - 3 = 0, (x + 1)² + y² = 4. Radius 2.

2. For the circle |z - 3| = 2|z|, find the real part of its centre.
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   -1. (x + 1)² + y² = 4 has centre (-1, 0).

3. z lies on the circle |z - 3| = 2|z|. Find the greatest value of |z|.
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   3. The centre is -1, at distance 1 from the origin; greatest |z| = 1 + 2 = 3, at z = -3. (Check: |-3 - 3| = 6 = 2|-3|.)

4. Which describes the locus arg((z - 2)/(z + 2)) = π/2?
   • The semicircle on the diameter from -2 to 2 with Im z < 0
   • The whole circle |z| = 2
   • The imaginary axis
   • The semicircle on the diameter from -2 to 2 with Im z > 0
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   The semicircle on the diameter from -2 to 2 with Im z > 0. The angle at z between the directions to 2 and to -2 is a right angle, so z is on the circle with diameter from -2 to 2. Testing z = 2i: (2i - 2)/(2i + 2) = i, whose argument is π/2. So it is the upper half.

5. The locus arg((z - 6)/(z - 2)) = π/4 is an arc of a circle. Find the radius of the circle, to 3 decimal places.
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   2.828. The chord from 2 to 6 has length 4 and subtends π/4 at the circumference, so π/2 at the centre. Radius = 4/(2sin(π/4)) = 2√2 = 2.828.

6. For the same locus arg((z - 6)/(z - 2)) = π/4, where is the centre of the circle?
   • 4 - 2i
   • 4
   • 4 + 2√2 i
   • 4 + 2i
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   4 + 2i. The centre is on the perpendicular bisector x = 4 at distance 2√2 from 2: (4 - 2)² + k² = 8, so k = ±2. The arc lies in the upper half-plane (test a point such as 4 + 4.83i), and for an angle of π/4 (acute) the centre is on the same side as the arc.

7. Why are the points a and b excluded from the locus arg((z - a)/(z - b)) = θ?
   • They lie inside the circle
   • The argument of 0, or of a quotient with zero denominator, is undefined
   • The locus is a straight line through them
   • They give an argument of π
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   The argument of 0, or of a quotient with zero denominator, is undefined. At z = a the top is 0, and at z = b the bottom is 0, so the argument is not defined there.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit F2
• Pearson Edexcel IAL Further Pure Mathematics 2 scheme of work

Last updated 8 November 2026. Spotted a mistake? Tell us.