Question

Rewrite in standard form. Find the center and radius of the circle.
(Equation below and medal goes to the best helper)

Answer 1

\[2x ^{2} + 2y ^{2} - 8x + 10y + 2 = 0\]

Answer 2

I have no idea how to do this :(

Answer 3

center is (2,-5/2)
radius is square root of 37 divided by 2

Answer 4

first step is to divide by 2

Answer 5

\[x^2+y^2-4x+5y+1=0\]

Answer 6

can you guys draw this out ??

Answer 7

next step is to complete the square twice
do you know how to complete the square?

Answer 8

complete the square ?? i dont believe so, i just started this lesson im a online student.

Answer 9

if you do not know how to compete the square, then you cannot do it

Answer 10

i can show you (maybe) it only takes a couple steps

Answer 11

ok maybe your using different terminology for something I do know~ and yea that would be helpful my lessons dont help much

Answer 12

ok so first off do you know what the goal is here?

Answer 13

to wright the equation in standard forum

Answer 14

find center and radius

Answer 15

right, which is \[(x-h)^2+(y-k)^2=r^2\] notice that \((x-h)^2\) and \(y-k)^2\) are "perfect squares" i.e. the square of something
that is why you have to "complete the square" twice
once for the \(x\) terms and once for the \(y\) terms

Answer 16

lets group them together first \[x^2-4x+y^2+5y=-1\]

Answer 17

oh thats what u meant never heard that term ~

Answer 18

so we have to fill in these blanks for \(h\) and \(k\) \[(x-h)^2+(y-k)^2=-1+something\]

Answer 19

what is half of 4?

Answer 20

2

Answer 21

and what is \(2^2\)?

Answer 22

4

Answer 23

so our start is \[(x-2)^2+(y-k)^2=-1+4=3\] now we repeat for the y term

Answer 24

what is half of 5?

Answer 25

2.5

Answer 26

yes, but don't use a decimal

Answer 27

so what do i use ? 2 ? 3?

Answer 28

no, a fraction

Answer 29

\[\frac{ 2 }{ 5 }\] ?

Answer 30

you got it upside down!!

Answer 31

oh soo \[\frac{ 5 }{ 2}\]

Answer 32

yeah
and what is \((\frac{5}{2})^2\)?

Answer 33

mmm gosh multiplying fractions :( ummm is it ... \[\frac{ 25 }{ 2 }\]

Answer 34

no, you forgot to square the two

Answer 35

ohh so \[\frac{ 25 }{ 4 }\]

Answer 36

I always thought you left the bottom number alone

Answer 37

right
so now we have \[(x-2)^2+(y+\frac{5}{2})^2=3+\frac{25}{4}\]

Answer 38

that is called "completing the square"
you have two competed squares on the left
almost done, last step is only to add on the right, then it is in standard form

Answer 39

you leave the bottom number alone if you are ADDING fractions with like denominators
for example \[\frac{5}{2}=\frac{8}{2}=\frac{13}{2}\]

Answer 40

but if you are multiplying you just multiply \[\left(\frac{5}{2}\right)^2=\frac{5}{2}\times \frac{5}{2}=\frac{25}{4}\]

Answer 41

ohh so this will be \[\left( x - 2 \right)^{2} +\left( y +\frac{ 5 }{ 2 } \right)^{2} = \frac{ 25 }{ 4} 3\]

Answer 42

?

Answer 43

what the heck kind of number is \(\frac{ 25 }{ 4} 3\)?? never seen a number that looked like that before

Answer 44

you gotta add !

Answer 45

well I thought i was adding I guess not ! ~ -.-

Answer 46

the way you wrote it, it looks like some sort of multiplication
not sure
in any case you get a fraction when you add

Answer 47

I added the 3 to the 25/4 at the end ~

Answer 48

\[3+\frac{25}{4}=\frac{3\times 4+25}{4}=\frac{37}{4}\]

Answer 49

oh I just added it as an whole number .. nvm alright soo i add that to the end and its the answer ?

Answer 50

ok well what about center and radius ?

Answer 51

\[(x-2)^2+(y+\frac{5}{2})^2=\frac{37}{4}\] is standard from
you read the center and radius from that form

Answer 52

the center is \[(2,-\frac{5}{2})\] and the radius is \[\frac{\sqrt{37}}{2}\]

Answer 53

ohhhh ~~~ how so they where already in the equation ?