If the Centre and Radius of the Circle |(z−2)/(z−3)| = 2 Are Respectively (α, β) and γ, Then 3(α + β + γ) Is Equal To

JEE Main 2023 — 1 February, Morning Shift. Previous Year Question.

Problem

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:

Key insight. A modulus equation like $|z-2|=2|z-3|$ isn’t a circle equation in the usual $(x-h)^2+(y-k)^2=r^2$ form — but substituting $z=x+iy$ and squaring both sides converts it directly into that form after expansion, revealing the general-form circle equation $x^2+y^2+2gx+2fy+c=0$ hiding inside.

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Approach

Rewrite the given modulus equation as $|z-2|=2|z-3|$, substitute $z=x+iy$, and square both sides to remove the moduli. Expand and simplify into the general circle form $x^2+y^2+2gx+2fy+c=0$, then read off the centre as $(-g,-f)$ and the radius as $\sqrt{g^2+f^2-c}$.

Solution

Step 1 — Rewrite the equation

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

Step 2 — Substitute z = x + iy

$$|(x-2)+iy| = 2|(x-3)+iy|$$

Step 3 — Square both sides

$$(x-2)^2+y^2 = 4\big[(x-3)^2+y^2\big]$$

Step 4 — Expand both sides

$$x^2-4x+4+y^2 = 4(x^2-6x+9+y^2) = 4x^2-24x+36+4y^2$$

Step 5 — Collect all terms on one side

$$x^2-4x+4+y^2-4x^2+24x-36-4y^2=0$$

$$-3x^2-3y^2+20x-32=0$$

Multiplying by $-1$ and dividing by $3$:

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

Step 6 — Match to the general circle form

Comparing with $x^2+y^2+2gx+2fy+c=0$:

$$2g=-\frac{20}{3} \implies g=-\frac{10}{3}, \qquad 2f=0 \implies f=0, \qquad c=\frac{32}{3}$$

Step 7 — Find the centre

$$(\alpha,\beta) = (-g,-f) = \left(\frac{10}{3},\ 0\right)$$

Step 8 — Find the radius

$$\gamma = \sqrt{g^2+f^2-c} = \sqrt{\frac{100}{9}+0-\frac{32}{3}} = \sqrt{\frac{100}{9}-\frac{96}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$$

Step 9 — Compute 3(α + β + γ)

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

Answer

Common mistakes

• Forgetting to square both sides before expanding. The equation $|z-2|=2|z-3|$ can’t be expanded term by term while moduli are still present — squaring first (using $|w|^2=w\bar{w}$, or equivalently the real-coordinate form) is what turns it into an ordinary algebraic equation.
• Misreading the centre formula. For $x^2+y^2+2gx+2fy+c=0$, the centre is $(-g,-f)$ — the negative of the half-coefficients, not $(g,f)$ directly.

Practise next

• If the centre and radius of the circle $\left|\dfrac{z-1}{z-4}\right|=3$ are $(\alpha,\beta)$ and $\gamma$, find $2(\alpha+\beta+\gamma)$, using the same substitute-and-expand method.