A point moves such that the sum of the squares of its distance from the sides of a square of side unity is equal to 9. Show that the locus is a circle whose centre coincides with the centre of the square. Find also its radius

Dear Debadutta

Let us take a simple square having four vertices at (.5,.5),(-.5,.5),(.5,-.5),(-.5,-.5),

Point is (x1,y1)

So according to question

(x1+.5)^2+(y1+.5)^2+(x1+.5)^2+(y1-.5)^2+(x1-.5)^2+(y1+.5)^2+(x1-.5)^2+(y1-.5)^2=9

4x1²+4y1²=9-8*.5²=7

So we can see locus is a circle whose centre coincides with the centre of the square and radius of circle is square root of 7/4.

All the best.

AKASH GOYAL

AskiitiansExpert-IITD

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