Question 1099803
<br>This is a curious type of problem that I have not seen before....<br>
Definitely try to get a second opinion....<br>
Let {{{z = a+bi}}}<br>
Then {{{(z-2i)/(2z-1) = (a+(b-2)i)/((2a-1)+(2b)i)}}}<br>
Rationalize the denominator by multiplying by its conjugate:
{{{((a+(b-2)i)((2a-1)-(2b)i))/(((2a-1)+(2b)i)((2a-1)-(2b)i)) = ((2a^2-2)+(-2abi)+((b-2)(2a-1)i)+(2b^2-4b))/((2a-1)+(2b)i)}}}<br>
The condition for that number to be purely imaginary is the the real part must be 0:
{{{2a^2-a+2b^2-4b = 0}}}<br>
With the a^2 and b^2 terms with the same coefficient, that is indeed the equation of a circle.<br>
To find the center and radius of the circle, complete the squares in a and b:
{{{2(a^2-(1/2)a+1/16) + 2(b^2-2b+1) = 0+2/16+2 = 17/8}}}
{{{(a-(1/4))^2 + (b-1)^2 = 17/16}}}<br>
The center of the circle is at ((1/4),1); the radius is sqrt(17)/4.