Rewrite in standard form. Find the center and radius of the circle.
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- anonymous

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- anonymous

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

- anonymous

I have no idea how to do this :(

- anonymous

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

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## More answers

- anonymous

first step is to divide by 2

- anonymous

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

- anonymous

can you guys draw this out ??

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

to wright the equation in standard forum

- anonymous

find center and radius

- anonymous

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

- anonymous

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

- anonymous

oh thats what u meant never heard that term ~

- anonymous

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

- anonymous

what is half of 4?

- anonymous

2

- anonymous

and what is \(2^2\)?

- anonymous

4

- anonymous

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

- anonymous

what is half of 5?

- anonymous

2.5

- anonymous

yes, but don't use a decimal

- anonymous

so what do i use ? 2 ? 3?

- anonymous

no, a fraction

- anonymous

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

- anonymous

you got it upside down!!

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

no, you forgot to square the two

- anonymous

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

- anonymous

I always thought you left the bottom number alone

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

?

- anonymous

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

- anonymous

you gotta add !

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ok well what about center and radius ?

- anonymous

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

- anonymous

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

- anonymous

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

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