Verified Solution • 2025–26 Session moderate

QJEEM23Ma3518: JEE Main 2023 Mathematics - Class 11 JEEM23Ma3518 Solution: Let A be the point ( 1 , 2 ) and B be any point on the curve x…

Let A be the point (1,2) and B be any point on the curve x2+y2=16. If the centre of the locus of the point P, which divides the line segment AB in the ratio 3:2 is the point C(α,β), then the length of the line segment AC is :

Multiple Choice Options

  • A 355
  • B 655
  • C 255
  • D 455

Step-by-Step Verified Solution

We have, equation of circle is x2+y2=16x^2+y^2=16

Let any point on the circle x2+y2=42x^2+y^2=4^2 is B(4cos⁡θ,4sin⁡θ)B(4 \cos \theta, 4 \sin \theta) and A(1,2)A(1,2)

Let P\mathrm{P} be (h,k)(h, k) which divides AB\mathrm{AB} in 3:23: 2

JEE Main 2023 (Online) 10th April Evening Shift Mathematics - Circle Question 33 English Explanation

So

h=12cos⁡θ+23+2 and k=12sin⁡θ+2×23+2

⇒cos⁡θ=5h−212\Rightarrow \cos \theta=\frac{5 h-2}{12} and sin⁡θ=5k−412\sin \theta=\frac{5 k-4}{12}

As, cos⁡2θ+sin⁡2θ=1,(5h−212)2+(5k−412)2=1\cos ^2 \theta+\sin ^2 \theta=1,\left(\frac{5 h-2}{12}\right)^2+\left(\frac{5 k-4}{12}\right)^2=1

⇒(h−25)2+(k−45)2=12252

∴\therefore Locus of point PP is (x−25)2+(y−45)2=12252\left(x-\frac{2}{5}\right)^2+\left(y-\frac{4}{5}\right)^2=\frac{12^2}{5^2}

which is equation of circle with centre (25,45)\left(\frac{2}{5}, \frac{4}{5}\right)

Hence, AC=(1−25)2+(2−45)2=455=355A C=\sqrt{\left(1-\frac{2}{5}\right)^2+\left(2-\frac{4}{5}\right)^2}=\frac{\sqrt{45}}{5}=\frac{3 \sqrt{5}}{5}

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