JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let circle \(C\) be the image of \(x^2+y^2-2 x+4 y-4=0\) in the line \(2 x-3 y+5=0\) and \(A\) be the point on \(C\) such that \(O A\) is parallel to \(x\)-axis and \(A\) lies on the right hand side of the centre \(O\) of \(C\). If \(B(\alpha, \beta)\), with \(\beta \lt 4\), lies on \(C\) such that the length of the are \(A B\) is \((1 / 6)^{\text {th }}\) of the perimeter of \(C\), then \(\beta-\sqrt{3} \alpha\) is equal to
- A \(3+\sqrt{3}\)
- B \(4\)
- C \(4-\sqrt{3}\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
Centre \((1,-2), r=3\) Reflection of \((1,-2)\) about \(2 x-3 y+5=0\) \(\begin{aligned} & \frac{x-1}{2}=\frac{y+2}{-3}=\frac{-2(2+6+5)}{13}=-2 \\ & x=-3, y=4 \end{aligned}\) Equation of circle ' C ' \(C:(x+3)^2+(y-4)^2=9\) A.T.Q.…
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