Question

An equation of a circle is x^2 + y^2 + 10x + 6y + 18 = 0.

Part 1: Show all your work in determining the center and radius of this circle. (3 points)

Part 2: In complete sentences, explain the procedure used. (3 points)

Answer 1

I have no clue how to do this one :(

Answer 2

\[x^2 + y2 + 10x + 6y + 18 = 0\] complete the square twice
first rewrite as
\[x^2+10x+y^2+6y=-18\]
do you know how to complete a square?

Answer 3

what is half of 10?

Answer 4

I don't remember... Can you remind me?

Answer 5

5

Answer 6

good, what is \(5^2\) ?

Answer 7

25

Answer 8

center is (-5,-3) and radius is sqrt(15)

Answer 9

ok so the first thing we are going to write instead of \[x^2+10x+y^2+6y=-18\] is
\[(x+5)^2+y^2+6y=-18+25\] or \[(x+5)^2+y^2+6y=7\]

Answer 10

now what is half of 6?

Answer 11

3

Answer 12

btw answer given above is wrong

Answer 13

good, okay, and again, what is \(3^2\) ?

Answer 14

?

Answer 15

9

Answer 16

@amistre64 can you help me finish this? @satellite73 left...

Answer 17

@ganeshie8

Answer 18

ok so the first thing we are going to write instead of

\(x^2+10x+y2+6y=-18\)

is

\((x+5)^2+y2+6y=-18+25\)

or

\((x+5)^2+y2+6y+9=7+9\)

Answer 19

ok so the first thing we are going to write instead of

\(x^2+10x+y^2+6y=-18\)

is

\((x+5)^2+y^2+6y=-18+25\)

or

\((x+5)^2+y^2+6y+9=7+9\)


or

\((x+5)^2+(y+3)^2=16\)

Answer 20

now, u knw the standard form of equation of circle wid center (h, k) and radius r ?

Answer 21

@ganeshie8 I got the center part, but what's the radius?

Answer 22

oh nice :) wat do u get for center ?

Answer 23

(-5,-3)

Answer 24

thats correct !

Answer 25

but how do I get the radius?

Answer 26

\((x-h)^2+(y-k)^2=r^2\)

center = \((h, k)\)
radius = \(r\)

Answer 27

so r=4?

Answer 28

just observe that 16 can be written as 4^2

Answer 29

Yes ! r = 4

Answer 30

Thanks soooooooooooooooo much!!!!

Answer 31

np :D

Answer 32

and plz medal satellite, not me :)