Question

Find the center of the circle with equation x^2+y^2-4x+6y+1=0

A) (-2,3)
B) (2,3)
C) (2,-3)
D) (-2,-3)

Answer 1

hey this is similar to the one yesterday xD just do the completing the square method

Answer 2

i just did it the way you tought me :D i got -2,3 is that right?

Answer 3

Its (2, -3)

Answer 4

you got the opposite signs :))

Answer 5

i forgot to tell that you should equate the ones inside the parentheses to 0

Answer 6

Also for this one Find the center of the ellipse with the equation 3x^2+4y^2+18x-32y=5 i got 3,-4

Answer 7

it's -3 ,4

Answer 8

can you show me the equation?

Answer 9

i keep getting switched up

Answer 10

for example you got
(x-2)^2 + (y+3)^2 =4

x-2 = 0
x=2

y+3=0
y=-3

Answer 11

3x^2+4y^2+18x-32y=5
(3x^2+18x)+(4y^2-32y)
(6x^2+8y^2)
3,-4

Answer 12

those are the steps i took ^^

Answer 13

oh don't group together the x and y they should be inside different parentheses

(3x^2+18x )+(4y^2-32y )=

Answer 14

oh ok

Answer 15

ok here's something that could also help you
notice that 3 and 18 can be are multiples of 3
you can take out the three to simplify the expressions

3(x^2+6x ) + 4(y^2-8y+ ) =

Answer 16

ok

Answer 17

then do the same as before

3(x^2+6x+9) + 4(y^2-8y+16) = 5 +27 + 64 ( i got 27 and 64 because i multiplied them to the number outside the parentheses respectively 3 and 4)

Answer 18

you can check out this website if you like to know more about equations on circles http://www.analyzemath.com/CircleEq/Tutorials.html

Answer 19

ok i got it :)