Locus of the equation

Question

One way to describe a set of points in the plane is by an equation or inequality in two variables, say $x$ and $y$. A solution of an equation in $x$ and $y$ is point $(x_0, y_0)$ in the plane for which the equation is true.

My questions: How do we describe a set points in the plane by an equation? Is there only single point which satisfies the equation? How do we know that the equation is true at a certain point? A circle is the locus of an equation: $(x - h)^2 + (y - k)^2 = r^2$ is this an equation in $x$ and $y$, I mean this seems to be an equation in change in x squared + change in y squared and $r^2$?

Answer 1

The Locus describes all the set of points through an equation.

The locus for any curve or figure is generalised by a condition which every point follows. Thus, we deduce an equation in concordance with the condition.

As per your example, the circle has the property that every point on its circumference will be at the same distance from the centre everywhere (known as radius).

Answer 2

In a set theoretic notation, we frequently use sets given by some (first order) properties, like $B:=\{a\in A\,\mid\, \varphi(a)$ holds$\}$.

In our case now $A=\Bbb R^2=\Bbb R\times\Bbb R=\{(x,y)\,\mid\, x,y\in\Bbb R\}$, the set of ordered pairs of real numbers, and $\varphi$ is an equation.

But, not only equations can describe (nice) subsets of the plane, e.g. $$H:=\{(x,y)\in\Bbb R^2\,\mid\,x\ge 0\}$$ gives the right half plane (which also includes the $y$-axis, i.e. '$x=0$').

In the given example of the circle (where $r,h,k$ are fixed), the crucial thing is that the sentence

The distance between $(x,y)$ and the given point $(h,k)$ is exactly $r$.

translates to an equation.

Answer 3

As the quote says, a set of points in the plane can be described by an equation involving two variables that we typically call $x$ and $y$. A specific point $(x_0,y_0)$ belongs to the set if the equation is satisfied when we put $x=x_0$ and $y=y_0$. So, take your circle example, for instance. Suppose we have a circle whose equation is $(x-1)^2 +(y-2)^2 = 25$. This is "an equation in $x$ and $y$", even though it includes some other stuff, too. Certainly, $x$ and $y$ are the only variables involved. There are many points (an infinite number) that satisfy this equation. Two obvious ones are $(4,6)$ and $(5,5)$, but there are infinitely many others. This is as we would expect -- a circle has an infinite number of points.

To find out whether the equation is true at a given point, we substitute the point's $(x,y)$ values into the equation, and see if we get a statement that's true. So, using the circle example again, we know that the point $(x,y)= (4,6)$ belongs to the circle because it is in fact true that $(4-1)^2 +(6-2)^2$ is equal to 25.