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The general equation for a circle is $$x^2+y^2+2gx+2fy+c = 0,$$ where $h = -g$ and $k = -f$. The radius is then $r =\sqrt{g^2+f^2-c}$.

The book says:

  1. If $g^2+f^2-c = 0$, then it's a point circle.
  2. If $g^2+f^2-c > 0$, then it's a real circle.
  3. If $g^2+f^2-c < 0$, then it's an unreal or imaginary circle.

What does that mean?

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    $\begingroup$ A point circle has radius zero and only includes a single point - the centre of the circle. A real/imaginary circle concerns whether the radius of the circle is real or imaginary. If the radius of a circle is imaginary then there are no real points on the circle. $\endgroup$ Jul 19, 2019 at 14:36

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A real circle is exactly what you normally think of as a circle; it has a radius that is a real number (not imaginary).

A point "circle" is just a point; it's a circle with a radius of zero (hence a degenerate circle).

An imaginary circle is one in which the radius is the square root of a negative number—i.e., imaginary.

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    $\begingroup$ Sir, can we call a point is Cartesian coordinates "a point circle" ? $\endgroup$
    – Ghost
    Jul 19, 2019 at 14:44
  • $\begingroup$ Sir, we are newbies.Can you please explain what can a circle look like if it's radius is zero? $\endgroup$
    – Ghost
    Jul 19, 2019 at 15:05
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    $\begingroup$ Who is "we"? And surely, you needn't call me "sir." Anyway, a circle is a point if its radius is zero. This seems to violate intuition regarding what a circle is, but keep in mind that a circle is merely the set of points that are a fixed distance away from the center. If that fixed distance is zero, then that set of points is simply the center point itself. In the same way, a line segment of length zero is a single point, a square of side length zero is a single point, a sphere of radius zero is a single point, and so on. $\endgroup$
    – Brian Tung
    Jul 19, 2019 at 16:54
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Short Answer

A circle can be described by the equation $$ r^2 = (x-h)^2 + (y-k)^2 $$ where $r$ is the radius of the circle and $(h,k)$ is the center of the circle.

  1. If $r^2 = 0$, then this equation has only one solution, the point $(h,k)$. Thus the equation describes a "point circle".

  2. If $r^2 > 0$, then this equation has many solutions of the form $(x,y)$, where both $x$ and $y$ are real numbers. The solutions form a circle in the real plane, thus the equation describes a "real circle".

  3. If $r^2 < 0$, then this equation has many solutions of the form $(x,y)$, but these solutions will be complex (not real). Because the equation is that for a circle, but the solutions are not real, we could describe this as an "imaginary circle" or an "unreal circle".

Explanation

Start from some basic intuition: a circle consists of all of the points in the plane which are of some constant distance (the radius) from some fixed point (the center). To fix notation, let $\mathscr{C}$ be the circle of radius $r$ centered at the point $(h,k)$. Again, the goal here is to build intuition, so we are going to start by assuming that $\mathscr{C}$ is a "normal" circle of the kind we encounter every day. As such, $(h,k)$ is an actual, honest-to-goodness point in the plane, which implies that $h$ and $k$ are both real numbers. More importantly, we are going to assume that the radius is a positive number—that is, $r>0$. A circle with zero, negative, or (heaven forbid!) complex radius just doesn't make sense to my intuition.

By definition, the circle $\mathscr{C}$ consists of all of the points $(x,y)$ which are a distance of $r$ units from the center $(h,k)$. This distance between a point $(x,y)$ and $(h,k)$ can be found using the distance formula: \begin{align} &r = d\bigl( (x,y), (h,k) \bigr) = \sqrt{ (x-h)^2 + (y-k)^2 } \\ &\qquad\implies \boxed{r^2 = (x-h)^2 + (y-k)^2} = x^2 - 2hx + h^2 + y^2 - 2ky + k^2 \\ &\qquad\implies x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - r^2 = 0. \end{align} Relabeling the variables gives the "general formula" for a circle given in the question, but the boxed question is (for my money) more helpful for understanding.

Once we get this formula, we can start playing with is "formally". That is, we can start relaxing our assumptions just to see what happens, without really caring about whether or not the results are meaningful or if they even make sense. Let's continue to assume that $(h,k)$ is a point in the real plane, but play around with $r$ (or, really, $r^2$).

As long as $r^2 > 0$, the equation $$ r^2 = (x-h)^2 + (y-k)^2 $$ will have solutions consisting of real-valued coordinate pairs $(x,y)$. These solutions form a real circle ("real solutions" / "real circle"). This handles case (2) from the originall question.

Now suppose that $r^2 = 0$ (this is case (1) from the original question). In this case, the circle consists of all of the points which satisfy the equation $$ 0^2 = (x-h)^2 + (y-k)^2. $$ But this equation has only one solution: $(x,y) = (h,k)$. Thus the "circle" described by this equation consists of only a single point, which we might reasonably call a "point circle".

Finally, suppose that $r^2 < 0$ (this is case (3) from the original question). In this case, we are looking for points $(x,y)$ such that $$ 0 > r^2 = (x-h)^2 + (y-k)^2. $$ There is no real number $r$ such that $r^2 < 0$, which means that any solutions to this equation must be imaginary (or complex, really). Such complex solutions do exist, and they satisfy the equation of a circle, so it is reasonable to refer to the set of solutions as an "imaginary circle" or "unreal circle".

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  • $\begingroup$ what's up with this late answer? $\endgroup$
    – Ghost
    Jul 13, 2020 at 18:05
  • $\begingroup$ The question was bumped by another late answerer (that answer has been deleted). I was unhappy with the accepted answer and the other (now deleted) answer, so I wrote one of my own. $\endgroup$
    – Xander Henderson
    Jul 13, 2020 at 18:14

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