Question

quick question :

how would i do this? [(X+3)/2]^2 + [(y-4)/2]^2 = 1

Answer 1

\[(\frac{ x+3 }{ 2 })^{2} +(\frac{ y-4 }{ 2 })^{2}=1\]

Answer 2

If it's the centre of the circle you want, it's (-3/2,2). If it's the radius, it's 1. I don't get the question in this. :\

Answer 3

i guess i just want to simplify it

Answer 4

i dont know what to do with the exponent?

Answer 5

Oh, eff the radius is 2.

Answer 6

It's a circle, open up the damn thing!

Answer 7

circles are technically ellipsi

Answer 8

ellipses**

Answer 9

okay, just help me simplify

Answer 10

i know what needs to be done, just help me distribute that exponent

Answer 11

\[x ^{2} + 6x+ 9 + y^{2} -8y + 16=4\]

Answer 12

okay break it down for me

Answer 13

what do i do first?

Answer 14

how do i distribute that exponent?

Answer 15

you know that property about [(a+b)^{2}]
Right?

Answer 16

but theres something right here it i woudl have to foil (a+b)(a+b)

Answer 17

It's the same thing.
It's happening in (x+3)^2 and (y-4)^2.

Answer 18

a^2+2ab+b^2

Answer 19

what about that 2 in the demominator?

Answer 20

I'm taking thw two from both the denominators out.

Answer 21

can you do that?

Answer 22

Mmhmm y'can. See, what I did there was,

(1/2)^{2}*(x+3)^2 + (1/2)^{2}*(y-4)^2=1

and then, after squaring (1/2) and taking it common in the LHS, I multiply both sides of the equation by 4.

Answer 23

Which gave me (x+3)^2+(y-4)^2=4

Answer 24

Hope that helped.

Answer 25

i dont know if i would just multiply 1/2 to both sides... dont i have to stick to order of operations?

Answer 26

exponents first...?

Answer 27

P.E.M.D.A.S?

Answer 28

You're making that complicated. Look, it's a valid term. You can simplify the exponents before solving the equation.

Answer 29

are you sure? i seems like i would be bound to use the exponent first

Answer 30

?

Answer 31

Even if we go by PEDMAS, the solution isn't different, just this approach is easier... and no problem, you can solve the exponent first.

\[\left( \frac{ x+3 }{ 2 } \right)^{2} +\left( \frac{ y-4 }{ 2 } \right)^{2} = 1\]

\[\left( \frac{ x^{2} }{ 4 } +\frac{ 6x }{ 4 }+ \frac{ 9 }{ 4 }\right) + \left( \frac{ y^{2} }{ 4 } -\frac{ 8y }{ 4 } +\frac{ 16 }{ 4 }\right)\\] = 1

Answer 32

well i'll be... i will trust you and just multiply by 2

Answer 33

why by 2 and not by 4 ?

Answer 34

(x+3)^2 (y-4)^2
------- + -------- = 1
2^2 2^2

take it on common denominator so for you need multiplie the right side by 4

than will get like result

(x+3)^2 +(y-4)^2 =4

ok ?

Answer 35

can you continue it now ?

Answer 36

I did it by multiplying by 4. -_-"