If the circle x square + y square +ax + 2y + 4=0. passes through (2,2) find the centre of the circle.

factor: x^2+y^2+ax+2y+4=0

First plug in the point (2,2) into the equation to find "a".

2^2+2^2+a(2)+2(2)+4=0

2a=-16

a=-8

So now with the value of a we have:

x^2+y^2-8x+2y+4=0

Let rearrange the terms:

x^2-8x+y^2+2y+4=0

Move the constant over to the right side of the equation.

x^2-8x+y^2+2y=-4

Completing the square is tricky but what needs to be done.

x^2-8x+16+y^2+2y+1=13

And then factor the sum and differences of squares.

(x-4)^2+(y+1)^2=13

Divide by 13.

(x-4)^2/13+(y+1)^2/13=0

Now it is in factored form, so we can get the center of the circle.

The center will be at (4,-1).

include a picture of the drawing in order for me to

step-by-step explanation: