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.
Best Answer
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.