give the center of circle with equation x^2+2x+y-10y+22=0?

Answer 1

x^2+2x+y-10y+22=0
is NOT a circle, there is a y^2 missing.
this is a parabola!!!! '

but if you meant
x^2+2x+y^2 -10y+22=0
then
x^2 + 2x +1 -1 + y^2 -10 y + 25 -25 + 22 =0
(x+1)^2 + (y-5)^2 =4

so center (-1,5)
radius: 2
'

Answer 2

I'm assuming you meant y² -10y, not y - 10y.

You always find the center of a circle by completing the square.

First, get all the variables on the left and numbers on the right:

x² + 2x + y² - 10y = -22

Then take the coefficient of x, 2, take 1/2 of it, 1, square it, 1, and add to both sides.

x² + 2x + 1 + y² - 10y = -22 + 1 = -21

Do the same with y: -10 is the coefficient of y, 1/2 of it is -5, squared is 25:

x² + 2x + 1 + y² - 10y + 25 = -21 + 25 = 4

Then unfoil the two perfect squares:

(x + 1)² + (y - 5)² = 4

And you'll see that you have a circle centered at (-1, 5) with radius √4 = 2.

Answer 3

Set x = 0. in case you get purely one attainable answer for x, then the y-axis touches once and is a tangent. x^2 + y^2 - 2x - 10y + 25 = 0 0^2 + y^2 - 2(0) - 10y + 25 = 0 y^2 - 10y + 25 = 0 are you able to imagine of two numbers that multiply to provide 25 and upload to -10? i love -5 and -5. sensible, it really is an similar huge style two times, yet that keeps to be 2 numbers. (y - 5)(y - 5) = 0 for this reason, y-5 = 0 y = 5 The circle touches the y-axis purely once at (0, 5). for this reason, the y-axis is a tangent to the circle.

Answer 4

Assuming you meant x^2+2x+y^2-10y+22=0

First factor:

(x^2+2x+1) + (y^2-10y+25) - 4 = 0
(x+1)^2 + (y-5)^2 = 4

The general formula for a circle is:
(x-h)^2+(y-k)^2=r^2 where the center is (h,k) and the radius if r.

in this case h=-1, k=5, and r=2

the center of the circle is (-1,5)

Answer 5

this is not the equation of a circle.
it is a parabola