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Circle question

2021 · 18 Mar · Shift 1 · Q33
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  5. /2021 · 18 Mar · Shift 1 · Q33

Circle question

2021 · 18 Mar · Shift 1 · Q33

JEE MainMathematicsCircleMCQ+4 / −1
For the four circles M, N, O and P, following four equations are given : Circle M : x2 + y2 = 1 Circle N : x2 + y2 −-− 2x = 0 Circle O : x2 + y2 −-− 2x −-− 2y + 1 = 0 Circle P : x2 + y2 −-− 2y = 0 If the centre of circle M is joined with centre of the circle N, further center of circle N is joined with centre of the circle O, centre of circle O is joined with the centre of circle P and lastly, centre of circle P is joined with centre of circle M, then these lines form the sides of a :
  1. A
    Rhombus
  2. B
    Square
  3. C
    Rectangle
  4. D
    Parallelogram
View written solutionFree

Correct answer: B

  1. Find the centres of the circles

A circle in the form x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 has centre (−g,−f).(-g,-f).(−g,−f).

Now rewrite each equation and identify its centre.

  • Circle M: x2+y2=1 ⇒ x2+y2−1=0x^2+y^2=1 \,\Rightarrow\, x^2+y^2-1=0x2+y2=1⇒x2+y2−1=0 So centre is M=(0,0).M=(0,0).M=(0,0).

  • Circle N: x2+y2−2x=0x^2+y^2-2x=0x2+y2−2x=0 Comparing with x2+y2+2gx+2fy+c=0,x^2+y^2+2gx+2fy+c=0,x2+y2+2gx+2fy+c=0, we get 2g=−2⇒g=−1,2f=0⇒f=0.2g=-2 \Rightarrow g=-1, \quad 2f=0 \Rightarrow f=0.2g=−2⇒g=−1,2f=0⇒f=0. Hence centre is N=(−g,−f)=(1,0).N=(-g,-f)=(1,0).N=(−g,−f)=(1,0).

  • Circle O: x2+y2−2x−2y+1=0x^2+y^2-2x-2y+1=0x2+y2−2x−2y+1=0 Here, 2g=−2⇒g=−1,2f=−2⇒f=−1.2g=-2 \Rightarrow g=-1, \quad 2f=-2 \Rightarrow f=-1.2g=−2⇒g=−1,2f=−2⇒f=−1. So centre is O=(1,1).O=(1,1).O=(1,1).

  • Circle P: x2+y2−2y=0x^2+y^2-2y=0x2+y2−2y=0 Here, 2g=0⇒g=0,2f=−2⇒f=−1.2g=0 \Rightarrow g=0, \quad 2f=-2 \Rightarrow f=-1.2g=0⇒g=0,2f=−2⇒f=−1. So centre is P=(0,1).P=(0,1).P=(0,1).

  1. Join the centres in the given order

The points are: M(0,0),  N(1,0),  O(1,1),  P(0,1).M(0,0),\; N(1,0),\; O(1,1),\; P(0,1).M(0,0),N(1,0),O(1,1),P(0,1).

These are the vertices of a quadrilateral formed by joining:

  • MNMNMN
  • NONONO
  • OPOPOP
  • PMPMPM
  1. Check side lengths

Using distance formula:

MN=(1−0)2+(0−0)2=1MN=\sqrt{(1-0)^2+(0-0)^2}=1MN=(1−0)2+(0−0)2​=1 NO=(1−1)2+(1−0)2=1NO=\sqrt{(1-1)^2+(1-0)^2}=1NO=(1−1)2+(1−0)2​=1 OP=(0−1)2+(1−1)2=1OP=\sqrt{(0-1)^2+(1-1)^2}=1OP=(0−1)2+(1−1)2​=1 PM=(0−0)2+(0−1)2=1PM=\sqrt{(0-0)^2+(0-1)^2}=1PM=(0−0)2+(0−1)2​=1

So all four sides are equal.

  1. Check angles
  • MNMNMN is horizontal.
  • NONONO is vertical.

Hence MN⊥NOMN \perp NOMN⊥NO, so one angle is 90∘90^\circ90∘.

A quadrilateral with all sides equal and one right angle is a square.

  1. Conclusion

The lines form the sides of a Square.\boxed{\text{Square}}.Square​.

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