Question

The centre of circle z = $\frac{3i - t}{2 + it}$ (tĪR) must be –

• A. $\left( 0,\frac{3}{4} \right)$
• B. (0, 0)
• C. $\left( 0,\frac{5}{4} \right)$
• D. None

Answer 1

Answer: (C)

$\left( 0,\frac{5}{4} \right)$

Answer 2

Sol. x + iy = $\frac{3i - t}{2 + it} \cdot \frac{2 - it}{2 - it}$ =$\frac{( - 2t + 3t) + i(6 + t^{2})}{4 + t^{2}}$

\ x = $\frac{t}{4 + t^{2}}$ & y = $\frac{6 + t^{2}}{4 + t^{2}}$

\ x² + y² = $\frac{t^{2} + 9}{t^{2} + 4}$ = $\frac{1 + 5}{t^{2} + 4}$ = 1+ $\frac{5}{2}$ (y – 1)

Ž x² + y² = $\frac{5y}{2} - \frac{3}{2}$ Žx² + y² – $\frac{5y}{2} + \frac{3}{2} = 0$