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Circle question

2023 · 10 Apr · Shift 1 · Q35
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  5. /2023 · 10 Apr · Shift 1 · Q35

Circle question

2023 · 10 Apr · Shift 1 · Q35

JEE MainMathematicsCircleMCQ+4 / −1
A line segment AB of length λ\lambdaλ moves such that the points A and B remain on the periphery of a circle of radius λ\lambdaλ. Then the locus of the point, that divides the line segment AB in the ratio 2 : 3, is a circle of radius :
  1. A
    23λ{2 \over 3}\lambda32​λ
  2. B
    35λ{3 \over 5}\lambda53​λ
  3. C
    197λ{{\sqrt {19} } \over 7}\lambda719​​λ
  4. D
    195λ{{\sqrt {19} } \over 5}\lambda519​​λ
View written solutionFree

Correct answer: D

  1. Set up the circle and chord

Let the fixed circle have center O(0,0)O(0,0)O(0,0) and radius λ\lambdaλ.

Since AAA and BBB lie on the circle and ABABAB has length λ\lambdaλ, the segment ABABAB is a chord of the circle of length λ\lambdaλ.

For a circle of radius λ\lambdaλ, if a chord subtends angle θ\thetaθ at the center, then

AB=2λsin⁡θ2.AB=2\lambda\sin\frac{\theta}{2}.AB=2λsin2θ​.

Given AB=λAB=\lambdaAB=λ, we get

2λsin⁡θ2=λ2\lambda\sin\frac{\theta}{2}=\lambda2λsin2θ​=λ ⇒sin⁡θ2=12\Rightarrow \sin\frac{\theta}{2}=\frac12⇒sin2θ​=21​ ⇒θ2=30∘⇒θ=60∘.\Rightarrow \frac{\theta}{2}=30^\circ \quad \Rightarrow \theta=60^\circ.⇒2θ​=30∘⇒θ=60∘.

So, for every position of the chord, the angle ∠AOB=60∘\angle AOB=60^\circ∠AOB=60∘.

  1. Choose coordinates for a general chord

Let the midpoint of chord ABABAB be MMM.

Because the chord subtends 60∘60^\circ60∘, the perpendicular distance of the chord from the center is

OM=λcos⁡30∘=32λ.OM=\lambda\cos 30^\circ=\frac{\sqrt3}{2}\lambda.OM=λcos30∘=23​​λ.

Also, half the chord length is

AM=BM=λ2.AM=BM=\frac{\lambda}{2}.AM=BM=2λ​.

Now take unit vectors:

  • n^\hat nn^ along OMOMOM,
  • t^\hat tt^ along the chord.

Then

M⃗=32λn^,\vec M=\frac{\sqrt3}{2}\lambda\hat n,M=23​​λn^,

\qquad \vec B=\vec M+\frac{\lambda}{2}\hat t.$$ --- 3. **Coordinates of the dividing point** Let $P$ divide $AB$ internally in the ratio $2:3$, i.e. $$AP:PB=2:3.$$ Using section formula in vector form, $$\vec P=\frac{3\vec A+2\vec B}{5}.$$ Substitute $\vec A,\vec B$: $$\vec P=\frac{3\left(\vec M-\frac{\lambda}{2}\hat t\right)+2\left(\vec M+\frac{\lambda}{2}\hat t\right)}{5}$$ $$=\frac{5\vec M-\frac{\lambda}{2}\hat t}{5}$$ $$=\vec M-\frac{\lambda}{10}\hat t.$$ So, $$\vec P=\frac{\sqrt3}{2}\lambda\hat n-\frac{\lambda}{10}\hat t.$$ Since $\hat n \perp \hat t$, the distance of $P$ from the center is constant: $$OP^2=\left(\frac{\sqrt3}{2}\lambda\right)^2+\left(\frac{\lambda}{10}\right)^2$$ $$=\frac{3}{4}\lambda^2+\frac{1}{100}\lambda^2 =\frac{75+1}{100}\lambda^2 =\frac{76}{100}\lambda^2 =\frac{19}{25}\lambda^2.$$ Hence, $$OP=\frac{\sqrt{19}}{5}\lambda.$$ Thus the locus is a circle centered at $O$ with radius $$\boxed{\frac{\sqrt{19}}{5}\lambda}. $$ --- 4. **Check options** - A: $\dfrac{2}{3}\lambda$ ❌ - B: $\dfrac{3}{5}\lambda$ ❌ - C: $\dfrac{\sqrt{19}}{7}\lambda$ ❌ - D: $\dfrac{\sqrt{19}}{5}\lambda$ ✅ So the correct option is **D**.
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