The center of the circle defined by the equation x squared - 4x + y squared = -28 is ____.?

Question

a. (-2, -5)
b. (-2, 5)
c. (2, 5)
d. (2, -5)

i need the solutions too. (: thaanks.

sorry. this is the real equation.

x squared - 4x t y squared -10y = -28

thanks!

Answer 1

*sigh*

I think your circle is:

x^2 - 4x + y^2 - 10y = -28

which you must convert into form

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

by completing the square

You do this by:

adding 4 to the x^2 - 4x bit (and also to the other side)

(x^2 - 4x + 4) + y^2 - 10y = -28 + 4

THEN you add 25 to the y^2 - 10y bit (and also to the other side)

(x^2 -4x + 4) + (y^2 - 10y + 25) = -28 + 4 + 25

then you simplify it all out ...

(x-2)^2 + (y-5)^2 = 1

and you have centre of circle at (2,5) ... is answer c

radius of 1

Answer 2

I am sure that the equation is like this

you may forget the +10y or -10y part
if it is +10y make it like this :
x^2 - 4x + y^2 +10y =-28
(x^2-4x+4) + (y^2+10y+25) =1
(x-2)^2+(x+5)^2=1, so the center is (2,-5) -> d
if it is -10ty make it like this :
x^2 - 4x + y^2 -10y =-28
(x^2-4x+4) + (y^2-10y+25) =1
(x-2)^2+(x+5)^2=1, so the center is (2,5) -> c

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Answer 3

I think you left out a -25y or +25 on the right-hand side of your equation.