JEE Mathematics 2023 Question with Solution

Given: $\left| \frac{z-2}{z-3} \right| = 2$

Find: The value of $3(\alpha+\beta+\gamma)$ where the center is $(\alpha,\beta)$ and radius is $\gamma$.

Let $z = x + iy$. Then

$$\sqrt{(x-2)^2 + y^2} = 2\sqrt{(x-3)^2 + y^2}$$

Squaring both sides,

$$(x-2)^2 + y^2 = 4\big((x-3)^2 + y^2\big)$$ $$x^2 + y^2 - 4x + 4 = 4x^2 + 4y^2 - 24x + 36$$ $$3x^2 + 3y^2 - 20x + 32 = 0$$

Dividing by $3$,

$$x^2 + y^2 - \frac{20}{3}x + \frac{32}{3} = 0$$

Comparing with the standard form of a circle,

$$\alpha = \frac{10}{3}, \qquad \beta = 0$$

Also,

$$\gamma = \sqrt{\left(\frac{10}{3}\right)^2 - \frac{32}{3}} = \sqrt{\frac{100}{9} - \frac{96}{9}} = \frac{2}{3}$$

Therefore,

$$3(\alpha + \beta + \gamma) = 3\left(\frac{10}{3} + 0 + \frac{2}{3}\right) = 12$$

The correct option is D. The first solution labels option A, but its working concludes the value is $12$, which matches option D.

Given: $\left| \frac{z-2}{z-3} \right| = 2$

Find: $3(\alpha+\beta+\gamma)$.

Use the direct distance form:

$$|z-2| = 2|z-3|$$

Substitute $z = x+iy$ and write distances from the points $(2,0)$ and $(3,0)$:

$$\sqrt{(x-2)^2+y^2} = 2\sqrt{(x-3)^2+y^2}$$

This immediately gives the circle

$$x^2 + y^2 - \frac{20}{3}x + \frac{32}{3} = 0$$

Hence center $\left(\frac{10}{3},0\right)$ and radius

$$\gamma = \sqrt{\left(\frac{10}{3}\right)^2 - \frac{32}{3}} = \frac{2}{3}$$

So,

$$3\left(\frac{10}{3}+0+\frac{2}{3}\right)=12$$

Thus the correct option is D.