1.

Show that the following equations represent a circle, and, find its centre and radius. (i) |z – 2 – i| = 3 (ii) |2z + 2 – 4i| = 2 (iii) |3z – 6 + 12i| = 8

Answer»

(i) Let z = x + iy 

|z – 2 – i| = 3 

⇒ |x + iy – 2 – i| = 3 

⇒ |(x – 2) + i(y – 1)| = 3

= √((x - 2)2 + (y -1)2) = 3

Squaring on both sides

(x – 2)2 + (y – 1)2 = 9 

⇒ x2 – 4x + 4 + y2 – 2y + 1 – 9 = 0 

⇒ x2 + y2 – 4x – 2y – 4 = 0 represents a circle 

2g = -4 ⇒ g = -2

2f = -2 ⇒ f = -1 

c = -4 

(a) Centre (-g, -f) = (2, 1) = 2 + i (b) 

(b) Radius = √(g2 + f2 - c) = √(4 + 1 + 4) = 3

Aliter: |z – (2 + i)| = 3 

Centre = 2 + i 

Radius = 3

(ii) |2(x + iy) + 2 – 4i| = 2 

⇒ |2x + i2y + 2 – 4i| = 2 

⇒ |(2x + 2) + i(2y – 4)| = 2 

⇒ |2(x + 1) + 2i(y – 2)| = 2 

⇒ |(x + 1) + i(y – 2)| = 1

= √((x + 1)2 + (y - 2)2) = 1

Squaring on both sides, 

x2 + 2x + 1 + y2 + 4 – 4y – 1 = 0

⇒ x2 + y2 + 2x – 4y + 4 = 0 represents a circle 

2g = 2 ⇒ g = 1 

2f = -4 ⇒ f = -2 

c = 4 

(a) Centre (-g, -f) = (-1, 2) = -1 + 2i 

(b) Radius = √(g2 + f2 - c) = √(1 + 4 - 4) = 1

Aliter: 2|(z + 1 – 2i)| 

= 2 |z – (-1 + 2i)| = 1 

Centre = -1 + 2i 

Radius = 1

(iii) |3(x + iy) – 6 + 12i| = 8 

⇒ |3x + i3y – 6 + 12i| = 8 

⇒ |3(x – 2) + i3 (y + 4)| = 8 

⇒ 3|(x – 2) + i (y + 4)| = 8

= 3√((x - 2)2 + (y + 4)2) = 8

Squaring on both sides, 9[(x – 2)2 + (y + 4)2] = 64 

⇒ x2 – 4x + 4 + y2 + 8y + 16 = 64/9

⇒ x2 + y2 – 4x + 8y + 20 – (64/9) = 0

x2 + y2 – 4x + 8y + (116/9) = 0 represents a circle. 

2g = -4 ⇒ g = -2 

2f = 8 ⇒ f = 4 

c = 116/9

(a) Centre (-g, -f) = (2, -4) = 2 – 4i 

(b) Radius = √(g2 + f2 - c) = √(4 + 16 - (116/9)) = √(180 - 116)/9 = 8/3

Aliter: |z – 2 + 4i| = 8/3

⇒ |z – (2 – 4i)| = 8/3

Centre = 2 – 4i, Radius = 8/3



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