How to factor this math problem??

Question

Here's the question:
Find the center and the radius for the circle x^2-2x+y^2-4y-4=0.

...somehow I have to "complete the square" and factor it...im not sure how to do it, can someone show like, a step-by-step method of solving this??
thanks!!!

Answer 1

1) Take everything without an x or y to the right.

x² - 2x + y² - 4y = 4

2) For each of the terms that has only x or only y:
2a) Divide the coefficient by 2. -2/2 = -1, -4/2 = -2.
2b) Square that number: (-1)² = 1, (-2)² = 4.
2c) Add those numbers to both sides.

x² - 2x + 1 + y² - 4y + 4 = 4 + 1 + 4

3) You now have two perfect squares for x and y on the left, and the radius squared on the right.

(x - 1)² + (y - 2)² = 9 = 3²

Since this is in the correct form:

(x - h)² + (y - k)² = r²

the answer is: center (1, 2), radius 3.

Answer 2

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

Now, the equation is (x-h)^2+(y-k)^2=r^2. Let's try to get to this form. First we add four to both sides to remove that -4. That gives

x^2-2x+y^2-4y=4. Next we "complete the square"

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

Adding 1 made x^2-2x+1, which is (x-1)^2. Adding another 4 made y^2-4y+4, which is (y-2)^2. This is why we call it "completing the square", we add numbers to make perfect squares of algebraic expressions. Now we have

(x-1)^2+(y-2)^2=9

This is in the form of (x-h)^2+(y-k)^2=r^2. Since 3^2=9, h=1, k=2, and r=3.

Therefore, the center is (1,2), and the radius is 3.

Answer 3

x^2 - 2x + y^2 - 4y - 4 = 0
x^2 - 2x + y^2 - 4y = 4
(x^2 - 2x) + (y^2 - 4y) = 4
(x^2 - 2x + 1 - 1) + (y^2 - 4y + 4 - 4) = 4
((x - 1)^2 - 1) + ((y - 2)^2 - 4) = 4
(x - 1)^2 - 1 + (y - 2)^2 - 4 = 4
(x - 1)^2 + (y - 2)^2 - 5 = 4
(x - 1)^2 + (y - 2)^2 = 9

The standard form of a circle is
(x - h)^2 + (y - k)^2 = r^2

whereas (h,k) is the center and "r" is the radius

ANS :
Center = (1,2)
Radius = 3

Answer 4

You have to complete 2 squares
x^2-2x+y^2-4y-4=0
(x^2-2x+1) +(y^2-4y+4) -4-1-4=0
(x-1)^2+(y-2)^2=9=3^2

The center is at (1, 2) & the radius is 3

Answer 5

x^2-2x + y^2 -4y -4= 0
(x^2-2x +1) + (y^2 - 4y + 4) = 9
(x-1)^2 + (y-2)^2 = 3^2

This is the equation of a circle with center C(1,2) and radius 3.

Answer 6

x^2-2x + y^2 -4y -4= 0
(x^2-2x +1) + (y^2 - 4y + 4) = 9
(x-1)^2 + (y-2)^2 = 3^2

the center is at (1,2) and the radius is 3.

Answer 7

a million. aspect thoroughly: x^2 + 10x + 25 A) (x + 5)(x - 5) B) (x + 5)(x + 5) C) (x - 5)(x - 5) D) (x + 25)(x + a million) answer is B. x*x =x^2; 5*5=25; 5*x+5*x =10x 2. aspect thoroughly: 3x^2 - 12 A) 3(x^2 - 4) B) 3(x + 2)(x + 2) C) 3(x + 2)(x - 2) D) (x + 2)(x - 2) answer is C. A is genuine yet did not go far sufficient. x^2-4 will be factored to (x+4)(x-4). 3. aspect thoroughly: 2x^2 - x - 10 A)(2x - 5)(x + 2) B) (2x + 5)(x - 2) C) (2x + 5)(x + 2) D) (2x - 5)(x - 2) answer is A. 2x*x =2x^2; (-5)(2)=-10;(-5)x +(2x)2 = -x 4. Simplify thoroughly: (4x -8) / (x-2) A) 4(x-2) / (x-2) B) 4 C) 8 D) 3x+4 answer is B. A did not cancel the x-2 words to get 4. it might that x won't be able to be equivalent to 2 or you need to be dividing by technique of 0 that is a no-no

Answer 8

you have to get in the form

(x-a)^2 +(y-b)^2 = c^2

now (x^2-2x) = x^2+2(x)(-1)+(-1)^2 -1 = (x-1)^2 -1

as (x-a)^2 = x^2 -2 a x + a^2

2ax = 2x so a =1 so add 1-1 when you add 1 it will complete the square and then subtract 1 for compasation

similarly

(y^2-4y) = (y-2)^2 - 4
we get (x-1)^2-1 + (y-2)^2 -4 -4 = 0
or (x-1)^2+(y-2)^2 = 9 = 3^2

so centre is (1,2) radius is 3