Question

If \((\alpha, \beta)\) is the external centre of similitude of the circles \[ x^2 + y^2 = 3 \] and \[ x^2 + y^2 - 2x + 4y + 4 = 0, \] then find \(\frac{\beta}{\alpha}\).

• A. \(-3\)
• B. \(-2\)
• C. \(2\)
• D. \(3\)

Hint:

Apply the external division formula for centres of similitude and simplify ratio carefully.

Answer 1

The Correct Option is B

First circle center \(C_1 = (0,0)\), radius \(r_1 = \sqrt{3}\). Rewrite second circle as \[ (x-1)^2 + (y+2)^2 = 1, \] so center \(C_2 = (1,-2)\), radius \(r_2 = 1\). External centre of similitude divides \(C_1 C_2\) externally in ratio \(r_1 : r_2 = \sqrt{3} : 1\). Using section formula, \[ \alpha = \frac{\sqrt{3} \times 1 - 1 \times 0}{\sqrt{3} - 1} = \frac{\sqrt{3}}{\sqrt{3} - 1},
\beta = \frac{\sqrt{3} \times (-2) - 1 \times 0}{\sqrt{3} - 1} = \frac{-2 \sqrt{3}}{\sqrt{3} - 1}. \] Therefore, \[ \frac{\beta}{\alpha} = \frac{-2 \sqrt{3}}{\sqrt{3}} = -2. \]