Question

A point $P$ moves in such a way that the ratio of its distance from two coplanar points is always a fixed number ( $\equiv 1$ ). Then identify the locus of the point.

Answer 1

Hint: A locus is a group of points with positions that meet the requirements for their locations and create geometric structures like lines, line segments, circles, curves, etc. Instead of just being a collection of points, the points can also be thought of as locations or points that can be moved.
The circle, in terms of the locus of points or loci, corresponds to all points that are equally spaced apart from one fixed point, with the fixed point being the circle's center and its radius being the collection of points outward from the center.

Complete step by step Solution:
Let two coplanar points $(0,0)\text{ }$ and $(a,0)$
Coplanar points are those that are located on the same plane, and non-coplanar points are those that are not located on the same plane. Any two points are collinear because we are aware that any two points in two dimensions can always cross a line. Similar to this, any three points in three-dimensional space can always pass across a plane, hence any three points are always coplanar.
distance between two point $=\lambda$
$\dfrac{\sqrt{(x-0)^{2}+(y-0)^{2}}}{\sqrt{(x-a)^{2}+y^{2}}}=\lambda$
squaring both sides
$\dfrac{\sqrt{x^{2}+y^{2}}}{\sqrt{(x-a)^{2}+y^{2}}}=\lambda$
$x^{2}+y^{2}+\left(\dfrac{\lambda^{2}}{\lambda^{2}-1}\right)\left(a^{2}-2 a x\right)=0$

Note:The general equation of a circle is another name for the centre of circle formula. If the radius is r, the center's coordinates are $(h,k),$and any point on the circle is$(x, y)$, the centre of the circle formula is as follows:
$(x-h)^{2}+(y-k)^{2}=r^{2}$
The equation for the centre of the circle is another name for this. In the sections that follow, we'll use this formula to determine a circle's equation or centre.