FP2 2016 5 Edexcel

Start by rearranging for z, we have:

w(z + 3) = wz + 3w = 2z - 1

So:

wz - 2z = z(w - 2) = -3w - 1

giving:

\displaystyle z = \frac {3w + 1} {2 - w}

The unit circle centred at the origin has equation |z| = 1, so:

\displaystyle \left|\frac {3w + 1} {2 - w}\right| = 1

so:

\left|3w + 1\right| = \left|2 - w\right|

Set w = u + iv, then:

\left|u + iv + 1\right| = \left|2 - u - iv\right|

equivalent to:

\left|3u + 3iv + 1\right|^2 = \left|2 - u - iv\right|^2

So:

(3u + 1)^2 + 3^2 v^2 = (2 - u)^2 + v^2

Expanding we have:

9u^2 + 6u + 1 + 9v^2 = u^2 - 4u + 4 + v^2

so:

8u^2 + 10u + 8 v^2 - 3 = 0

dividing by 8 we have:

\displaystyle \left(u^2 + \frac 5 4 u\right) + v^2 - \frac 3 8 = 0

completing the square we have:

\displaystyle \left(u + \frac 5 8\right)^2 - \left(\frac 5 8\right)^2 + v^2 - \frac 3 8 = 0

so:

\displaystyle \left(u + \frac 5 8\right)^2 + v^2 = \frac {49} {64}

So the centre of C is \displaystyle \left(-\frac 5 8, 0\right) and the radius is \displaystyle \sqrt {\frac {49} {64}} = \frac 7 8.

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