Verified Solution • 2025–26 Session moderate

QJEEM25Ma2c9c: JEE Main 2025 Mathematics - Class 11 JEEM25Ma2c9c Solution: Let circle be the image of in the line and be the point on such …

Let circle CC be the image of x2+y2−2x+4y−4=0x^2+y^2-2 x+4 y-4=0 in the line 2x−3y+5=02 x-3 y+5=0 and AA be the point on CC such that OAO A is parallel to xx -axis and AA lies on the right hand side of the centre OO of CC . If B(α,β)B(\alpha, \beta) , with \beta<4 , lies on CC such that the length of the arc ABA B is (1/6)th (1 / 6)^{\text {th }} of the perimeter of CC , then β−3α\beta-\sqrt{3} \alpha is equal to

Multiple Choice Options

  • A 4−34-\sqrt{3}
  • B 33
  • C 44
  • D 3+33+\sqrt{3}

Step-by-Step Verified Solution

JEE Main 2025 (Online) 24th January Morning Shift Mathematics - Circle Question 2 English Explanation 1

Centre (1,−2),r=3(1,-2), r=3

Reflection of (1,−2)(1,-2) about 2x−3y+5=02 x-3 y+5=0

amp;x−12=y+2−3=−2(2+6+5)13=−2amp;x=−3,y=4\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 ' CC '

C:(x+3)2+(y−4)2=9C:(x+3)^2+(y-4)^2=9

A.T.Q.

JEE Main 2025 (Online) 24th January Morning Shift Mathematics - Circle Question 2 English Explanation 2

amp;ℓ(arcAB⁡)=16×2πramp;rθ=16×2πramp;θ=π3amp;(α+6)2+(β−4)2=27amp;(α+3)2±(β−4)2=9(α+6)2−(α+3)2=18amp;⇒6α=−9amp;⇒α=−32,β=(4−332)amp;∴β−3αamp;(4−332)+332amp;=4\begin{aligned}& \ell(\operatorname{arcAB})=\frac{1}{6} \times 2 \pi \mathrm{r} \\& \mathrm{r} \theta=\frac{1}{6} \times 2 \pi \mathrm{r} \\& \theta=\frac{\pi}{3} \\& (\alpha+6)^2+(\beta-4)^2=27 \\& \frac{(\alpha+3)^2 \pm(\beta-4)^2=9}{(\alpha+6)^2-(\alpha+3)^2=18} \\& \Rightarrow 6 \alpha=-9 \\& \Rightarrow \alpha=\frac{-3}{2}, \beta=\left(4-\frac{3 \sqrt{3}}{2}\right) \\& \therefore \beta-\sqrt{3} \alpha \\& \left(4-\frac{3 \sqrt{3}}{2}\right)+\frac{3 \sqrt{3}}{2} \\& =4\end{aligned}

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