Question

How do I graph a circle with this equation:
(x^2)+(y^2)-2x+4y-20=0

Answer 1

do you know what a "perfect square trinomial" is?

Answer 2

Sort of

Answer 3

I know what it is but I don't know how to turn it into that

Answer 4

Please help

Answer 5

well, let's start by grouping
\(\bf x^2+y^2-2x+4y-20=0 \implies (x^2-2x)+(y^2+4y)-20=0\quad \\\quad \\
(x^2-2x+\square?)+(y^2+4y+\square?)-20=0\)

what values do you think we need in each group, to get a perfect square trinomial?
what's the missing value in each?

Answer 6

-8 and -12 maybe? I started grouping already

Answer 7

But I included the -20 in mine

Answer 8

hmm right.. so... \(\bf (x^2-2(\square)x+\square^2)+(y^2+2(\square)y+\square^2)-20=0\)

what do you think we need now?

Answer 9

4 and 6

Answer 10

Does that sound right?

Answer 11

out middle term in a perfect square trinomial
is the product of the other 2 terms, without the square

Answer 12

http://www.statisticslectures.com/images/trinomial11.gif

Answer 13

so, can't be 4 or 6, in either case

Answer 14

product of the other 2 terms, without the square * 2 , that is

Answer 15

But that can't be

Answer 16

Could you show me what you would do and explain how you got it?

Answer 17

I am getting more confused

Answer 18

lemme do the 1st one...
\(\bf (x^2-2x+\square^2)\\
2 \times x \times \square = 2x\\
\square = 1\\
(x^2-2x+\square^2) \implies (x^2-2x+1^2) \implies (x-1)^2\)

Answer 19

Is the second one (y+1)^2?

Answer 20

\(\bf \large{ (y^2+4y+\square^2)\\
2 \times y \times \square = 4y}\)

Answer 21

well... not quite 1

Answer 22

2

Answer 23

(y-2)^2

Answer 24

?

Answer 25

\(\bf (y^2+4y+\square^2)\\
2 \times y \times \square = 4y\\
\square = 2\\
(y^2+4y+\square^2)\implies (y^2+4y+2^2) \implies (y+2)^2\)

Answer 26

Right

Answer 27

So it should be (x-1)^2 + (y+2)^2 = 20?

Answer 28

notice though, we added 1 first, for one, and then 4 for the other

what we are doing is just borrowing from zero, so if we ADD 1 and 4, we have to also subtract 1 and 4

so
\(\bf x^2+y^2-2x+4y-20=0 \implies (x^2-2x)+(y^2+4y)-20=0\quad \\\quad \\
(x^2-2x+1^2)+(y^2+4y+2^2) - 1^2-2^2-20=0\\\quad \\

(x-1)^2+(y+2)^2- 25=0 \implies(x-1)^2+(y+2)^2 = 25\)