TS EAMCET · Maths · Circle
The centre of the circle \(r^2-4 r(\cos \theta+\sin \theta)-4=0\) in cartesian coordinates is
- A \((1,1)\)
- B \((-1,-1)\)
- C \((2,2)\)
- D \((-2,-2)\)
Answer & Solution
Correct Answer
(C) \((2,2)\)
Step-by-step Solution
Detailed explanation
Put \(\quad x=r \cos \theta\) and \(y=r \sin \theta\) \(\therefore \quad r^2=x^2+y^2\) From Eqs. (i)…
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