Obtain the Cartesian equation for the locus of z = x + iy in each of t

1.	Obtain the Cartesian equation for the locus of z = x + iy in each of the following cases. (i) \|z – 4\| = 16 (ii) \|z – 4\|2 – \|z – 1\|2 = 16
Answer» (i) z = x + iy \|z – 4\| = 16 ⇒ \|x + iy – 4\| = 16 ⇒ \|(x – 4) + iy\| = 16 = √((x - 4)²+ y²) = 16 Squaring on both sides (x – 4)² + y² = 256 ⇒ x² – 8x + 16 + y² – 256 = 0 ⇒ x² + y² – 8x – 240 = 0 represents the equation of circle (ii) \|x + iy – 4\|² – \|x + iy – 1\|² = 16 ⇒ \|(x – 4) + iy\|² – \|(x – 1) + iy\|² = 16 ⇒ [(x – 4)² + y²] – [(x – 1)² + y²] = 16 ⇒ (x² – 8x + 16 + y²) – (x² – 2x + 1 + y²) = 16 ⇒ x² + y² – 8x + 16 – x² + 2x – 1 – y² = 16 ⇒ -6x + 15 = 16 ⇒ 6x + 1 = 0

Obtain the Cartesian equation for the locus of z = x + iy in each of the following cases. (i) |z – 4| = 16 (ii) |z – 4|2 – |z – 1|2 = 16

Answer»

(i) z = x + iy

|z – 4| = 16

⇒ |x + iy – 4| = 16

⇒ |(x – 4) + iy| = 16

= √((x - 4)²+ y²) = 16

Squaring on both sides

(x – 4)² + y² = 256

⇒ x² – 8x + 16 + y² – 256 = 0

⇒ x² + y² – 8x – 240 = 0 represents the equation of circle

(ii) |x + iy – 4|² – |x + iy – 1|² = 16

⇒ |(x – 4) + iy|² – |(x – 1) + iy|² = 16

⇒ [(x – 4)² + y²] – [(x – 1)² + y²] = 16

⇒ (x² – 8x + 16 + y²) – (x² – 2x + 1 + y²) = 16

⇒ x² + y² – 8x + 16 – x² + 2x – 1 – y² = 16

⇒ -6x + 15 = 16

⇒ 6x + 1 = 0