Question

find the center and radius of the given circle. x^2 +y^2+6x-8y-39=0

Answer 1

Write it like this?
(x^2+6x + )+(y^2-8y + )=39

Answer 2

Now complete the square for the x part and again or the y part. Don't forget to add the numbers to both sides!

Answer 3

@Mertsj how do I complete the square?

Answer 4

Divide the coefficient of x by 2. What do you get?

Answer 5

for x^2-6x it would be 3 and for y^2-8y it would be 4 right?

Answer 6

Now square the 3 and the 4 and add those numbers to both sides.

Answer 7

(x^2+6x+9)+(y^2-8y+16)=39+9+16

Answer 8

Now factor the two trinomials.

Answer 9

why did I divide by two? common factor?

Answer 10

Because the relationship between the coefficient of x and the constant term in a trinomial square is always: ( 1/2 the coefficient of x )^2= the constant term.

Answer 11

\[x^2-6x+9=(x-3)^2 and (\frac{6}{2})^2=9\]

Answer 12

ok so I have (x-3)^2+(x+4)^2 = 8^2

Answer 13

\[x^2+10x+25=(x+5)^2 and (\frac{10}{2})^2=25\]

Answer 14

yep

Answer 15

Oh wait. I think it is( y-4)^2

Answer 16

\[(x+3)^2+(y-4)^2=64\]

Answer 17

So the center is????

Answer 18

(-3,4) r=8

Answer 19

yep

Answer 20

Good job!!

Answer 21

thanks. I like to be walked though these when I don't know them well.

Answer 22

yw

Answer 23

wait why was it x-4^2?

Answer 24

It is (y-4)^2 because the factors of y^2-8y+16 are (y-4)^2

Answer 25

0
(x^2+6x+9)+(y^2-8y+16)=39+9+16

Answer 26

ok. gotcha

Answer 27

Good.

Answer 28

@Mertsj can you walk me through one more to make sure I understand the concept?

Answer 29

Yep

Answer 30

If you post it.