Mathematical 'language' (geometry)

Question

What does this question mean:

'Show (translated from my native language) that the equation $ x^2 - 4x + y^2 + 6y = 51 $ is a circle.' I have absolutely no idea how to 'show/prove/etc.' it, other than plotting it/drawing it, which probably isn't what is meant.

Answer 1

Remember that the general form for a circle in the Cartesian co-ordinate system is:

$$(x-a)^{2}+(y-b)^{2}=r^{2},\tag{1}$$

Where the center of the circle is at $(a,b)$ and the radius is $r$.

Therefore, for your example $x^{2}-4x+y^{2}+6y=51$, we must show that it can be reduced to this form to show it is a circle.

We begin by completing the square for $x$, to give us:

$$x^{2}-4x=(x-2)^{2}-4,$$

Then, similarly for $y$:

$$y^{2}+6y=(y+3)^{2}-9,$$

Putting this in our original form, we have:

$$(x-2)^{2}-4+(y+3)^{2}-9=51$$

Re-arranging, we get:

$$(x-2)^{2}+(y+3)^{2}=64$$

Which by $(1)$, is a circle with center $(2,-3)$ and radius $\sqrt{64}=8$.

Hope this helps!

Answer 2

The correct formulation would be "Show that the set of points $(x,y) \in \mathbb{R}^2$ satisfing the equation is a circle." By definition, a circle is the set of points that have the same distance to a fixed origin (the center). So to prove that those points satisfing the equation constitute a circle, you could for example show that these are the same points that satisfy an equation of the form $r^2 = (x - x_c)^2 + (y - y_c)^2$ for values $r, x_c, y_c$ which you would have to find out.

In conclusion, you should try to manipulate the equation via equivalence relations into the standard form above (hint: completing the square).

Answer 3

Maybe you could rewrite the equation in the form $$(x-h)^2 + (y-k)^2 = r^2.$$ Now, a circle is just a set of points at a fixed distance (say $r$) from a fixed point (like $(h,k)$). So, if we think about the distance formula $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2},$$ we can interpret the given equation as.....