Question:
<p>A point P moves such that the sum of the squares of its distances from the sides of a given square is a constant. Then point P moves on</p>
<p style="display:inline">a circle with centre as the centre of the square</p>
<p style="display:inline">a circle passing through the centre of the square</p>
<p style="display:inline">a circle with centre as one of the vertices of the square</p>
<p style="display:inline">a straight line</p>
Step-by-Step Solution
Key Concept: By placing the square's sides along the coordinate axes, the geometric sum of squared distances converts into a quadratic equation in x and y that defines a circle.
<html><body><p><img alt="" data-imgur-src="3XcXTts.png" height="125" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/1623761654-jzhjwg.jpg" width="123"/><br/>
Let ‘a’ be the side length of the square and the moving point P be (h, k).<br/>
Let the square be placed such that one vertex is at the origin and two adjacent sides lie along the coordinate axes. Then according to the given condition,<br/>
h<sup>2</sup> + (h - a)<sup>2</sup> + k<sup>2</sup> + (k - a)<sup>2</sup> = constant = c<sup>2</sup> (say)<br/>
<span class="math-tex">$\Leftrightarrow$</span> h<sup>2</sup> + k<sup>2</sup> - a(h + k) + a<sup>2</sup> - <span class="math-tex">$\frac{c^{2}}{2}$</span> = 0<br/>
<span class="math-tex">$\Rightarrow$</span> (h, k) satisfies x<sup>2</sup> + y<sup>2</sup> - xa - ya + a<sup>2</sup> -<span class="math-tex">$\frac{c^{2}}{2}$</span> = 0<br/>
<span class="math-tex">$\Leftrightarrow\left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{a}{2}\right)^{2}=\frac{c^{2}-a^{2}}{2}$</span> ...(i)<br/>
Equation (i) represents a circle with centre at <span class="math-tex">$\left(\frac{a}{2}, \frac{a}{2}\right)$</span>, which is same as the centre of the square.</p></body></html>
Correct Answer: A