Can you help me solve this circle?

Question

x^2+y^2-3x-4y=0

Answer 1

You first need to put it in this form: (x-a)^2 + (y-b)^2 = r^2 which is a circle centered at (a,b) with a radius of r; in this case...
(x^2-3x) + (y^2-4y) = 0
I have down this block before, so you have to complete the square of both the LHS terms to get it in circle form.
(x^2-3x+9/4) - 9/4 + (y^2 - 4y + 4) - 4 = 0
where I have added and subtracted 9/4 and 4 to complete the square in each one:
(x-3/2)^2 + (x-2)^2 = 4 + 9/4 = 16/4 + 9/4 = 25/4
Sooooo...we have a circle centered at (3/2,2) with radius of sqr[25/4] = 5/2

Answer 2

You have to complete the square twice, one for the x, one for the y. First, rearrange the terms so the x's and y's are next to eachother.
x^2-3x+y^2-4y=0
To complete the square, add numbers to both sides of the equation, so the x^2 and y^2 terms are perfect squares.

x^2-3x+___+y^2-4y+___=0+__+__
(Take the middle number, divide by 2, then square it. This will always result in a perfect square.)

x^2-3x+9/4+y^2-4y+4=9/4+4=25/4

Now factor the x terms and y terms and you'll have your equation for a circle.

(x-3/2)^2+(y-2)^2=25/4

It is a circle of radius 5/2 centered at (3/2,2)

Answer 3

x^2 + y^2 - 3x - 4y = 0
x^2 - 3x + y^2 - 4y = 0
(x^2 - 3x) + (y^2 - 4y) = 0
(x^2 - 3x + (9/4) - (9/4)) + (y^2 - 4y + 4 - 4) = 0
((x - (3/2))^2 - (9/4)) + ((y - 2)^2 - 4) = 0
(x - (3/2))^2 - (9/4) + (y - 2)^2 - 4 = 0
(x - (3/2))^2 + (y - 2)^2 - (25/4) = 0
(x - (3/2))^2 + (y - 2)^2 = (25/4)

Center : ((3/2),2)
Radius : (5/2)

Answer 4

x²+y²-3x-4y=0
x²-3x+9/4+y²-4y+4=0+9/4+4
(x-3/2)²+(y-2)²=25/4

center= 3/2, 2
radius=5/2

Answer 5

Will take a guess and assume that you are asking for centre and radius?
Centre (3/2 , 2)
r = √(9/4 + 4 ) = √(25/4) = 5 / 2

Answer 6

Waht is the question?