Question

The sum of the square of the distance of a point from $(0,0),(0,1),(1,1),(1,0)$ is $18$ and its locus is a circle.

If $\mathrm{d}$ is the diameter of the circle, then find ${\mathrm{d}}^2$?

Answer 1

Compute the required value.

Let $P\left(x,y\right)$ be the point.

Given : The sum of the square of the distance of a point from $(0,0),(0,1),(1,1),(1,0)$ is $18$.

Let's plot the given points.

Thus, ${\left(\mathrm{OP}\right)}^2+{\left(\mathrm{AP}\right)}^2+{\left(\mathrm{PC}\right)}^2+{\left(\mathrm{PB}\right)}^2=18$

$\Rightarrow$ $x^2+y^2+x^2+{(y-1)}^2+{(x-1)}^2+y^2+{(x-1)}^2+{(y-1)}^2=18$

$\Rightarrow$$x^2+y^2+x^2+y^2+1-2y+x^2+1-2x+y^2+x^2+1-2x+y^2+1-2y=18$

$\Rightarrow$ $4x^2+4y^2-4x-4y+4=18$

$\Rightarrow$ $x^2+y^2-x-y-\frac{7}{2}=0$ …..$\left(i\right)$

We know that for the standard equation of the circle $x^2+y^2+2gx+2fy+c=0$ radius is given by $r=\sqrt{g^2+f^2-c}$

Now comparing equation $\left(i\right)$ with the standard form of the circle

$g=-\frac{1}{2},f=-\frac{1}{2},c=-\frac{7}{2}$

So, $r=\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{7}{2}}$

$\Rightarrow$ $r=2$

$\Rightarrow$ $d=2r$

$\Rightarrow$ $\mathrm{d}=4$

$\Rightarrow$${\mathrm{d}}^2=16$

Hence the value of ${\mathrm{d}}^2$ is $16$.