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Circle question

2021 · 26 Feb · Shift 2 · Q27
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  5. /2021 · 26 Feb · Shift 2 · Q27

Circle question

2021 · 26 Feb · Shift 2 · Q27

JEE MainMathematicsCircleMCQ+4 / −1
Let A(1, 4) and B(1, −-− 5) be two points. Let P be a point on the circle (x −-− 1)2 + (y −-− 1)2 = 1 such that (PA)2 + (PB)2 have maximum value, then the points, P, A and B lie on :
  1. A
    a straight line
  2. B
    an ellipse
  3. C
    a parabola
  4. D
    a hyperbola
View written solutionFree

Correct answer: A

  1. Given data
  • A(1,4)A(1,4)A(1,4)
  • B(1,−5)B(1,-5)B(1,−5)
  • P(x,y)P(x,y)P(x,y) lies on the circle (x−1)2+(y−1)2=1(x-1)^2+(y-1)^2=1(x−1)2+(y−1)2=1

We need to maximize PA2+PB2PA^2+PB^2PA2+PB2 and then determine what kind of curve the points P,A,BP,A,BP,A,B lie on.

  1. Write PA2PA^2PA2 and PB2PB^2PB2

Since P=(x,y)P=(x,y)P=(x,y), PA2=(x−1)2+(y−4)2PA^2=(x-1)^2+(y-4)^2PA2=(x−1)2+(y−4)2 PB2=(x−1)2+(y+5)2PB^2=(x-1)^2+(y+5)^2PB2=(x−1)2+(y+5)2

Therefore, PA2+PB2=2(x−1)2+(y−4)2+(y+5)2PA^2+PB^2=2(x-1)^2+(y-4)^2+(y+5)^2PA2+PB2=2(x−1)2+(y−4)2+(y+5)2

Expand: (y−4)2=y2−8y+16(y-4)^2=y^2-8y+16(y−4)2=y2−8y+16 (y+5)2=y2+10y+25(y+5)^2=y^2+10y+25(y+5)2=y2+10y+25

So, PA2+PB2=2(x−1)2+2y2+2y+41PA^2+PB^2=2(x-1)^2+2y^2+2y+41PA2+PB2=2(x−1)2+2y2+2y+41

  1. Use the circle condition

From (x−1)2+(y−1)2=1(x-1)^2+(y-1)^2=1(x−1)2+(y−1)2=1 we get (x−1)2+y2−2y+1=1(x-1)^2+y^2-2y+1=1(x−1)2+y2−2y+1=1 (x−1)2+y2=2y(x-1)^2+y^2=2y(x−1)2+y2=2y

Now substitute into the expression: PA^2+PB^2=2ig((x-1)^2+y^2ig)+2y+41 =2(2y)+2y+41=2(2y)+2y+41=2(2y)+2y+41 =6y+41=6y+41=6y+41

Thus, maximizing PA2+PB2PA^2+PB^2PA2+PB2 is equivalent to maximizing yyy.

  1. Find the maximum possible yyy on the circle

The circle (x−1)2+(y−1)2=1(x-1)^2+(y-1)^2=1(x−1)2+(y−1)2=1 is centered at (1,1)(1,1)(1,1) with radius 111.

Hence the maximum value of yyy is y=1+1=2y=1+1=2y=1+1=2 which occurs at the topmost point of the circle: P=(1,2)P=(1,2)P=(1,2)

  1. Check the positions of A,B,PA,B,PA,B,P

We have:

  • A=(1,4)A=(1,4)A=(1,4)
  • B=(1,−5)B=(1,-5)B=(1,−5)
  • P=(1,2)P=(1,2)P=(1,2)

All three points have the same xxx-coordinate x=1x=1x=1.

Therefore, they are collinear and lie on the straight line x=1x=1x=1

  1. Option check
  • A: a straight line — Correct
  • B: an ellipse — Incorrect
  • C: a parabola — Incorrect
  • D: a hyperbola — Incorrect
  1. Final answer

The points P,A,BP,A,BP,A,B lie on a straight line.

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