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AP EAMCET · Maths · Circle

Let \(\alpha, \beta\) be the roots of \(x^2+5 x+6=0\) and \(\gamma, \delta\) be the roots of \(y^2+6 y+7=0\). Then the equation of the circle with \((\alpha, \gamma)\) and \((\beta, \delta)\) as the extremities of a diameter is

  1. A \(x^2+y^2+5 x+6 y+10=0\)
  2. B \(x^2+y^2+5 x+6 y+11=0\)
  3. C \(x^2+y^2+5 x+6 y+13=0\)
  4. D \(x^2+y^2+5 x+6 y+12=0\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(x^2+y^2+5 x+6 y+13=0\)

Step-by-step Solution

Detailed explanation

\((x - \alpha)(x - \beta) + (y - \gamma)(y - \delta) = 0\) \(x^2 - (\alpha+\beta)x + \alpha\beta + y^2 - (\gamma+\delta)y + \gamma\delta = 0\) \(x^2 - (-5)x + 6 + y^2 - (-6)y + 7 = 0\) \(x^2 + y^2 + 5x + 6y + 13 = 0\)
From AP EAMCET
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