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Circle question

2021 · 26 Aug · Shift 1 · Q39
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  5. /2021 · 26 Aug · Shift 1 · Q39

Circle question

2021 · 26 Aug · Shift 1 · Q39

JEE MainMathematicsCircleNumerical+4 / −1
The locus of a point, which moves such that the sum of squares of its distances from the points (0, 0), (1, 0), (0, 1), (1, 1) is 18 units, is a circle of diameter d. Then d2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 16

  1. Let the moving point be P(x,y)P(x,y)P(x,y).

  2. The given condition is: PA2+PB2+PC2+PD2=18PA^2+PB^2+PC^2+PD^2=18PA2+PB2+PC2+PD2=18 where A(0,0),  B(1,0),  C(0,1),  D(1,1).A(0,0),\; B(1,0),\; C(0,1),\; D(1,1).A(0,0),B(1,0),C(0,1),D(1,1).

  3. Compute each squared distance: PA2=x2+y2PA^2=x^2+y^2PA2=x2+y2 PB2=(x−1)2+y2PB^2=(x-1)^2+y^2PB2=(x−1)2+y2 PC2=x2+(y−1)2PC^2=x^2+(y-1)^2PC2=x2+(y−1)2 PD2=(x−1)2+(y−1)2PD^2=(x-1)^2+(y-1)^2PD2=(x−1)2+(y−1)2

  4. Add them: [ (x^2+y^2)+((x-1)^2+y^2)+(x^2+(y-1)^2)+((x-1)^2+(y-1)^2)=18 ]

Expand: [ x^2+y^2+(x^2-2x+1+y^2)+(x^2+y^2-2y+1)+(x^2-2x+1+y^2-2y+1)=18 ]

Combine like terms: 4x2+4y2−4x−4y+4=184x^2+4y^2-4x-4y+4=184x2+4y2−4x−4y+4=18

Divide by 444: x2+y2−x−y+1=92x^2+y^2-x-y+1=\frac{9}{2}x2+y2−x−y+1=29​ x2+y2−x−y=72x^2+y^2-x-y=\frac{7}{2}x2+y2−x−y=27​

  1. Complete squares: x2−x=(x−12)2−14x^2-x=\left(x-\frac12\right)^2-\frac14x2−x=(x−21​)2−41​ y2−y=(y−12)2−14y^2-y=\left(y-\frac12\right)^2-\frac14y2−y=(y−21​)2−41​

So, [ \left(x-\frac12\right)^2-\frac14+\left(y-\frac12\right)^2-\frac14=\frac72 ] [ \left(x-\frac12\right)^2+\left(y-\frac12\right)^2=\frac72+\frac12=4 ]

Thus the locus is a circle with radius r=2r=2r=2 So diameter d=2r=4d=2r=4d=2r=4 Hence, d2=16d^2=16d2=16

  1. Comparison with stored answer: Stored correct answer = 161616, which matches our result.
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