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JEE Main 202118 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
18 March
Shift
Evening Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2021 · 18 March · Evening Shift · Mathematics · Circle

Let S _ 1 : x ^ 2 + y ^ 2 = 9 and S _ 2 : ( x - 2 ) ^ 2 + y ^ 2 = 1...

Let S1:x2+y2=9{S}_{1}: {x}^{2} + {y}^{2} = 9 and S2:(x−2)2+y2=1{S}_{2}: ( x - 2{)}^{2} + {y}^{2} = 1.

Then the locus of center of a variable circle SS which touches S1{S}_{1} internally and S2{S}_{2} externally always passes through the points :

Let S _ 1 : x ^ 2 + y ^ 2 = 9 and S _ 2 : ( x - 2 ) ^ 2 + y ^ 2 = 1 . Then the locus of center of a variable circle SLet S _ 1 : x ^ 2 + y ^ 2 = 9 and S _ 2 : ( x - 2 ) ^ 2 + y ^ 2 = 1 . Then the locus of center of a variable circle SLet S _ 1 : x ^ 2 + y ^ 2 = 9 and S _ 2 : ( x - 2 ) ^ 2 + y ^ 2 = 1 . Then the locus of center of a variable circle S

Options

  1. A(0,±3)( 0, \pm \sqrt{3} )
  2. B(12,±52)\left(\dfrac{1}{2}, \pm \dfrac{\sqrt{5}}{2}\right)
  3. C(2,±32)\left(2, \pm \dfrac{3}{2}\right)
  4. D(1,±2)( 1, \pm 2 )

Correct answer

C. (2,±32)\left(2, \pm \dfrac{3}{2}\right)

Step-by-step solution

Let S _ 1 : x ^ 2 + y ^ 2 = 9 and S _ 2 : ( x - 2 ) ^ 2 + y ^ 2 = 1 . Then the locus of center of a variable circle S

∵c1c2=r1−r2∵ {c}_{1}{c}_{2} = {r}_{1} - {r}_{2}

Let S _ 1 : x ^ 2 + y ^ 2 = 9 and S _ 2 : ( x - 2 ) ^ 2 + y ^ 2 = 1 . Then the locus of center of a variable circle S

∴∴ Given circles are touching internally

Let a variable circle with centre PP and radius rr

⇒PA=r1−randPB=r2+r⇒ PA = {r}_{1} - r and PB = {r}_{2} + r

⇒PA+PB=r1+r2⇒ PA + PB = {r}_{1} + {r}_{2}

⇒PA+PB=4(>AB)⇒ PA + PB = 4 ( > AB )

⇒⇒ Locus of PP is an ellipse with foci at A(0,0)andB(2,0)A ( 0, 0 ) and B ( 2, 0 ) and length of major axis is 2a=42a = 4, e=12e = \dfrac{1}{2}

⇒⇒ Centre is at ( 1, 0 )andb2=a2(1−e2)=3and {b}^{2} = {a}^{2}\left(1 - {e}^{2}\right) = 3

if xx-axis is the axis of ellipse then

Let S _ 1 : x ^ 2 + y ^ 2 = 9 and S _ 2 : ( x - 2 ) ^ 2 + y ^ 2 = 1 . Then the locus of center of a variable circle S

⇒E:(x−1)24+y23=1⇒ E: \dfrac{( x - 1{)}^{2}}{4} + \dfrac{{y}^{2}}{3} = 1

which is satisfied by \left(2, \pm \dfrac{3}{2}\right)

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