U"X\Xb2+Y\Xb2+4x-4y-28=0 To Standard Form Then Give The Coordinates Of The Center And The Radius ( I Hope Yall Help Me)"

x²+y²+4x-4y-28=0 to standard form then give the coordinates of the center and the radius ( i hope yall help me)

Answer:

The coordinates of the center of the circle is (-2,2) while its radius is 6.

Step-by-step explanation:

Step 1: Write the given

$x^{2} +y^{2}+4x-4y-28=0$
The equation is an equation for circle since the coefficient and sign of $x^{2}$ and $y^{2}$ is equal.

Step 2: Find the center-radius form of the given as well as the center coordinates and the radius of the circle.

Center-radius form: $(x-h)^{2} +(y-k)^{2} =radius^{2}$
Where (h,k) will be the coordinates of the center of the circle
In order to transform the given equation into the center-radius form then we will use the technique "completing the square"
Group the terms with the same variable and isolate the constant: $(x^{2} +4x)+(y^{2}-4y)=28$
In completing the square, we need to work both on x and y variable separately.
Divide the coefficient of 4x by 2 and square the resulting answer: $(\frac{4}{2})^{2} =4$
Divide teh coefficient of -4y by 2 and square the resulting answer: $(\frac{-4}{2})^{2} =4$
The equation will result to: $(x^{2} + 4x+4)+(y^{2} -4y+4)=28+4+4$
4+4 is also added to the constant in order for the equation to remain the same.
Convert the group of terms of x an y variable to squared form: $(x+2)^{2} +(y-2)^{2}=36$
The center of the circle is (-2,2)
Since $radius^{2} =36$ , then radius=6.