If the centre and radius of the circle |z-2z-3|=2 are respectively (,) and , then 3(++) is…

Question

If the centre and radius of the circle $\left|\dfrac{z-2}{z-3}\right|=2$ are respectively $(\alpha,\beta)$ and $\gamma$, then $3(\alpha+\beta+\gamma)$ is equal to

Accepted Answer

$$(x-2)^2+y^2=4\big[(x-3)^2+y^2\big] \implies x^2+y^2-4x+4=4x^2+4y^2-24x+36$$
$$\implies 3x^2+3y^2-20x+32=0 \implies x^2+y^2-\frac{20}3x+\frac{32}3=0.$$

Centre $(\alpha,\beta)=\left(\dfrac{10}3,0\right)$, and $\gamma=\sqrt{\left(\dfrac{10}3\right)^2-\dfrac{32}3}=\sqrt{\dfrac{100-96}9}=\sqrt{\dfrac49}=\dfrac23$.

$$3(\alpha+\beta+\gamma)=3\left(\frac{10}3+0+\frac23\right)=3\cdot4=12.$$

Solutions · 1