The Student Room Group
I may be wrong, but:

Distance from centre of circle to P = 2. Using a^2 + b^2 = c^2 (where c is the line from the centre to P), a^2 and b^2 are the same, so 2a^2 = 4, a^2 = 2 and a = root 2.

So the point P would be (4 - root 2), (2 + root 2)j
Start from O

Need to go 4 + 2j to get to the centre of the circle

Now to get to P from circle centre, have to move radial distance (i.e. 2)

Can see from diagram that point P sits at exactly 45 degrees from the purely real axis (i.e. a horizontal line)

use tanθ=imagreal tan \theta = \frac{imag}{real}

implies

imagreal=tan(45)=1 \frac{|imag|}{|real|} = tan(45) = 1

therefore

imag=real |imag| = |real|

Now use pythagoras theorem

imag2+real2=radius2=22=4 |imag|^2 + |real|^2 = radius^2 = 2^2 = 4

So combining the two we get

imag2+imag2=4 |imag|^2 + |imag|^2 = 4

implies

imag=real=2 |imag| = |real| = \sqrt{2}

Full answer therefore

P = 42+(2+2)j 4 - \sqrt{2} + (2 + \sqrt{2})j

hope that makes some sense!

Latest