WBJEE · Maths · Circle
If one end of a diameter of the circle \(3 x^{2}+3 y^{2}-9 x+6 y+5=0\) is \((1,2),\) then the other end is
- A (2,1)
- B (2,4)
- C (2,-4)
- D (-4,2)
Answer & Solution
Correct Answer
(C) (2,-4)
Step-by-step Solution
Detailed explanation
Given equation of circle is \[ \begin{array}{r} 3 x^{2}+3 y^{2}-9 x+6 y+5=0 \\ \Rightarrow \quad x^{2}+y^{2}-3 x+2 y+\frac{5}{3}=0 \end{array} Centre =\left(\frac{3}{2},-1\right) \] and radius \(=\sqrt{\frac{9}{4}+1-\frac{5}{3}}\)…
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