Question

The center of the arc $arg\left(\frac{z-6-3i}{z-8-6i}\right)=\frac{\pi }{4}$ is

Answer 1

Given that:-

$arg\left(\frac{z-6-3i}{z-8-6i}\right)=\frac{\pi }{4}$

To find:- Centre of the arc $arg\left(\frac{z-6-3i}{z-8-6i}\right)=\frac{\pi }{4}$

Solution:-

Let $z=x+iy$

$\therefore \left(\frac{z-6-3i}{z-8-6i}\right)=\frac{(x+iy)-6-3i}{(x+iy)-8-6i}$

$\Rightarrow =\frac{(x-6)+i(y-3)}{(x-8)+i(y-6)}$

On multiplying and dividing with the conjugate of denominator, we have

$\left(\frac{z-6-3i}{z-8-6i}\right)=\frac{(x-6)+i(y-3)}{(x-8)+i(y-6)}\times \frac{(x-8)-i(y-6)}{(x-8)-i(y-6)}$

$\Rightarrow =\frac{(x^2+y^2-14x-9y+66)+i(3x-2y-12)}{{(x-8)}^2+{(y-6)}^2}$

$\Rightarrow \left(\frac{z-6-3i}{z-8-6i}\right)=\frac{(x^2+y^2-14x-9y+66)}{{(x-8)}^2+{(y-6)}^2}+i\frac{3x-2y-12}{{(x-8)}^2+{(y-6)}^2}=a+ib\left(say\right)$

Given that:-

$arg\left(\frac{z-6-3i}{z-8-6i}\right)=\frac{pi}{4}$

$\therefore \frac{b}{a}=\tan \frac{\pi }{4}$

$\Rightarrow \frac{b}{a}=1\Rightarrow a=b$

$\therefore \frac{(x^2+y^2-14x-9y+66)}{{(x-8)}^2+{(y-6)}^2}=\frac{3x-2y-12}{{(x-8)}^2+{(y-6)}^2}$

$\Rightarrow x^2+y^2-14x-9y+66=3x-2y-12$

$\Rightarrow x^2+y^2-14x-9y+66-3x+2y+12=0$

$\Rightarrow x^2+y^2-17x-7y+78=0.....\left(1\right)$

Equation $\left(1\right)$ is an equation of circle

Therefore comparing it with the general equation of circle, i.e., $x^2+y^2+2gx+2fy+c=0$, we get

$g=-\frac{17}{2}$

$f=-\frac{7}{2}$

$\therefore$ Centre of circle will be $(-g,-f)=(\frac{17}{2},\frac{7}{2})$.