Multiple choice

Let two circles with radii $r_1$ and $r_2$ have their centers at a distance of $ \sqrt{2} \left(r_1-r_2 \right)$. If $r_1$ and $r_2$ are roots of the equation $x^2 - 2(\alpha + \beta)x + \alpha^2 + \beta^2=0$, then the relation between $\alpha$ and $\beta$ for which the two circles are orthogonal is

  1. $\alpha^{ 2} + \beta^{2} =4\alpha \beta $
  2. $(\alpha +\beta)^{2}=4\alpha \beta $
  3. $\alpha ^{2}+\beta ^{2}=\alpha \beta $
  4. $\alpha ^{2}-\beta ^{2}=4\alpha \beta$
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A Correct answer
Explanation

For orthogonal circles, 2*g1*g2 + 2*f1*f2 = c1 + c2. Given the distance between centers is sqrt(2)(r1-r2), the condition simplifies to the relation between radii and centers. The roots of x^2 - 2(a+b)x + a^2+b^2 = 0 are r1, r2. The condition for orthogonality leads to a^2 + b^2 = 4ab.

AI explanation

The sum of the roots is 2 times the sum of alpha and beta, and the product of the roots is alpha squared plus beta squared. For the circles to be orthogonal, the square of the distance between their centers must equal the sum of the squares of their radii, so 2 times the quantity r1 minus r2 squared equals r1 squared plus r2 squared. Expanding and substituting the root relationships gives 2 times the quantity 4 times alpha times beta minus the sum of alpha squared and beta squared equaling alpha squared plus beta squared, which simplifies to alpha squared plus beta squared equals 4 times alpha times beta. The result is alpha squared plus beta squared equals 4 alpha beta.

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