show that the equation represents a circle and find the centre

Question

I have the following equation:

${x^2 + y^2 - 4x -6y + 9 = 0}$

And I am asked to show that the equation represents a circle and find the centre. In order to show that the equation represents a circle is it enough to get it into the ${(x - h)^2 + (y -k)^2 = r^2}$ format.

I tried that and used splitting the square to say:

${x^2 -4 x + 4 + y^2 -6y + 9 = -9 + 4 + 9}$

Which then factors as

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

I would have said this proves the equation is a circle because it meets the format

But the answer in the text book is

(2, 3) and radius 2, I see where it gets this from the equation but I don't understand how it proves the equation is a circle.

Answer 1

If I understand correctly your question is why $(x-h)^2+(y-k)^2=r^2$ represents a circle. Simple. Take (positive) sqrt on both sides, then equation says distance between all points $(x,y)$ from a fixed point $(h,k)$ is $r$. This clearly indicates figure is a circle.

Answer 2

Hint:

the equation

$$ x^2+y^2+ax+by+c=0 $$

represents a circle of center $C=(\alpha,\beta)=(-a/2,-b/2)$ iff $\alpha^2+\beta^2 -c=r^2 >0$ , and $r$ is the radius.

Inthis case the equation can be write as $$ (x-\alpha)^2+(y-\beta)^2 = r^2 $$

and this means that

$$ \sqrt{(x-\alpha)^2+(y-\beta)^2}=r $$

where the Left side is the distance of a point $P=(x,y)$ from the center $C=(\alpha,\beta)$.

Answer 3

The equation of any circle has the following three characteristics:-

(0). It is an equation of degree $2$;

(1). (($x^2$)) = (($y^2$)); and

(2). (($xy$)) = 0

You just have to check whether the given equation meets the above requirements or not. It is absolutely not necessary to change it to center-radius form before you can claim that represents a circle, unless this is an elementary exercise given by your teacher as a training.

Furthermore, there are ways to “read” off the center and the radius directly from the given equation (which has been given in “standard form” already). That is, there is no need to make such a conversion, unless this is another training.

In case of an imaginary or a degenerated circle is not accepted, we need to add the following extra condition to test it:_

(3) (($x$)$)^2 + $ (($y$)$)^2 - 4 \cdot ($($x^0$) $) \gt 0$