TS EAMCET · Maths · Circle
The centres of all circles passing through the points of intersection of the circles \(x^2+y^2+2 x-2 y+1=0\) and \(x^2+y^2-2 x+2 y-2=0\) and having radius \(\sqrt{14}\) lie on the curve
- A \(x+y=0\)
- B \(y^2=4 x-2\)
- C \(3 x^2+5 x=y\)
- D \(2 x^2+3 y^2=7\)
Answer & Solution
Correct Answer
(A) \(x+y=0\)
Step-by-step Solution
Detailed explanation
\((x^2+y^2+2 x-2 y+1) + \lambda (x^2+y^2-2 x+2 y-2) = 0\) \((1+\lambda)x^2 + (1+\lambda)y^2 + (2-2\lambda)x + (-2+2\lambda)y + (1-2\lambda) = 0\) \(x^2 + y^2 + \frac{2(1-\lambda)}{1+\lambda}x + \frac{2(\lambda-1)}{1+\lambda}y + \frac{1-2\lambda}{1+\lambda} = 0\) Center…
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