Question

Match the circle equations in general form with their corresponding equations in standard form.
Tiles

x2 + y2 − 4x + 12y − 20 = 0

(x − 6)2 + (y − 4)2 = 56

x2 + y2 + 6x − 8y − 10 = 0

(x − 2)2 + (y + 6)2 = 60

3x2 + 3y2 + 12x + 18y − 15 = 0

(x + 2)2 + (y + 3)2 = 18

5x2 + 5y2 − 10x + 20y − 30 = 0

(x + 1)2 + (y − 6)2 = 46

2x2 + 2y2 − 24x − 16y − 8 = 0

x2 + y2 + 2x − 12y − 9 = 0

Answer 1

@Hero can you help with this tutorial?

Answer 2

Okay, so starting with the first one
\(x^2 + y^2 − 4x + 12y − 20 = 0\)

First begin by pair x terms together and y terms together like so:
\(x^2 - 4x + y^2 + 12y - 20 = 0\)

Answer 3

Next, add 20 to both sides:
\(x^2 - 4x + y^2 + 12y = 20\)

Answer 4

So now we have two binomial expressions on the left side \(x^2 - 4x\) and \(y^2 + 12y\)

Notice that -4 and 12 play the role of the coefficient b in the expression \(ax^2 + bx + c\) so we can add \(\left(\dfrac{b}{2}\right)^2\) to both sides for each binomial expression like so:

Answer 5

\(x^2 - 4x + (-2)^2 + y^2 + 12y + (6)^2 = 20 + (-2)^2 + (6)^2\)

Answer 6

@BeautifulAngel are you following so far?

Answer 7

Yes, just writing all this down

Answer 8

The previous step above is knowing as completing the square, so if you are not familiar with that, there are plenty videos on it on youtube.

The previous step simplifies to
\(x^2 - 4x + 4 + y^2 + 12x + 36 = 20 + 4 + 36\)

or just
\(x^2 - 4x + 4 + y^2 + 12x + 36 = 60\)

You can write each trinomial square on the left side in the form \(\left(x + \dfrac{b}{2}\right)^2\) like so
\((x - 2)^2 + (y + 6)^2 = 60\)

Answer 9

Which should match with one of your listed pairings.

Answer 10

If I did solved (x−2)2+(y+6)2=60
would it give me the other pair as well?

Answer 11

Yes, you can reverse the process to get the other pair.

Answer 12

Or rather, I should say you can get the other pair by expanding \((x - 2)^2 + (y + 6)^2 = 60\)

Answer 13

Would it be a more difficult process?

Answer 14

You should do whichever is most convenient for you. Either way is the same level of difficulty. I don't think one way is more convenient than the other. I prefer to build than deconstruct. Whichever method you choose is a matter of preference.

Answer 15

Ok. Thank you for helping @Hero :)

Answer 16

You're welcome.