Two points on a ground are 1 m apart. If a cow moves in the field in such a way that it's distance from the two points is always in ratio 3: 2 then

Let us consider the two points which are 1 m apart to be$$A(0,0)$$ and $$B(1,0)$$

Let at any instant the position of cow be denoted by coordinates $$C(x,y)$$

So, AC=$$\sqrt{\ x^2+y^2}$$

BC=$$\sqrt{\left(\ x-1\right)^2+y^2}$$

Given,$$\dfrac{AC}{BC}=\dfrac{3}{2}$$

or, $$\dfrac{AC^2}{BC^2}=\dfrac{9}{4}$$

or, $$\dfrac{x^2+y^2}{\left(x-1\right)^2+y^2}=\dfrac{9}{4}$$

or, $$4\left(x^2+y^2\right)=9\left(\left(x-1\right)^2+y^2\right)$$

or, $$4x^2+4y^2=9x^2-18x+9+9y^2$$

or, $$5x^2+5y^2-18x+9=0$$

This a second degree equation in $$x$$ and $$y$$ with coefficients of $$x$$ and $$y$$ being equal.

Therefore, this equation represents a circle.

Hence, we can conclude the cow moves in a circle.