TS EAMCET · Maths · Circle
The equation of the circle which passes through the origin and cuts orthogonally each of the circles \(x^2+y^2-6 x+8=0\) and \(x^2+y^2-2 x-2 y=7\) is
- A \(3 x^2+3 y^2-8 x-13 y=0\)
- B \(3 x^2+3 y^2+8 x+29 y=0\)
- C \(3 x^2+3 y^2+8 x+29 y=0\)
- D \(3 x^2+3 y^2-8 x-29 y=0\)
Answer & Solution
Correct Answer
(B) \(3 x^2+3 y^2+8 x+29 y=0\)
Step-by-step Solution
Detailed explanation
Let the required equation of circle be \(x^2+y^2+2 g x+2 f y=0\). Since, the above circle cuts the given circles orthogonally.…
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