TS EAMCET · Maths · Circle
If the origin lies on a diameter of the circle \(x^2+y^2-4 x-2 y-4=0\), then the equation of the circle passing through the end points of that diameter and the point \((1,2)\) is
- A \(x^2+y^2-2 x-4 y=0\)
- B \(3 x^2+3 y^2-19 x+8 y-12=0\)
- C \(7 x^2+7 y^2-31 x-28 y+17=0\)
- D \(x^2+y^2=5\)
Answer & Solution
Correct Answer
(B) \(3 x^2+3 y^2-19 x+8 y-12=0\)
Step-by-step Solution
Detailed explanation
Given circle \(x^2+y^2-4 x-2 y-4=0\) Centre \((2,1)\) Equation of diameter of circle passes through origin. \(x-2 y=0\) Equation of circle through the end point of diameter of circle and line \(x-2 y=0\) is \(x^2+y^2-4 x-2 y-4+\lambda(x-2 y)=0\) Since, this is passes through…
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