AP EAMCET · Maths · Circle
The circles \(x^2+y^2+2 x+3 y-7=0\) and \(x^2+y^2+4 x-7 y+5=0\) intersect at the points \(\mathrm{A}\) and \(\mathrm{B}\). The equation of the circle, having \(\overline{\mathrm{AB}}\) as a diameter is
- A \(26 x^2+26 y^2+77 x-47 y+32=0\)
- B \(26 x^2+26 y^2+77 x+47 y-32=0\)
- C \(26 x^2+26 y^2+77 x-47 y-32=0\)
- D \(26 x^2+26 y^2+77 x+47 y+32=0\)
Answer & Solution
Correct Answer
(C) \(26 x^2+26 y^2+77 x-47 y-32=0\)
Step-by-step Solution
Detailed explanation
\(S_1 = x^2+y^2+2 x+3 y-7=0\) \(S_2 = x^2+y^2+4 x-7 y+5=0\) Equation of circle through intersection of \(S_1\) and \(S_2\): \(S_1 + \lambda S_2 = 0\) \((1+\lambda)x^2+(1+\lambda)y^2+(2+4\lambda)x+(3-7\lambda)y+(-7+5\lambda)=0\) Center of this circle…
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