Question

What is the general form of the equation for the given circle?


A.x2 + y2 − 8x − 8y + 23 = 0



B.x2 + y2 − 8x − 8y + 32 = 0



C.x2 + y2 − 4x − 4y + 23 = 0



D.x2 + y2 + 4x + 4y + 9 = 0

Answer 1

Hello,
Answer A

B: x²+y²-8x-8y+32=0==>(x-4)²+(y-4)²=0 is a point, not a circle.
C:x²+y²-4x-y+23=0==>(x+2)²+(y-2)²=-15 is not a circle
D: x²+y²+4x+4y+9=0==>(x+2)²+(y+2)²=-1 is not a circle.
But;x²+y²-8x-8y+23=0==>(x-4)²+(y-4)=9 is a circle.

Answer 2

Answer:

The answer is A. $x^2+y^2-8x-8y+23=0$

Step-by-step explanation:

In order to determine the general form of the equation, first we have to know the standard form of the circle.

The standard form is:

$(x-h)^2+(y-k)^2=r^2$

Where

(h,k)= coordinates of the center of the circle

r= radius of the circle

So, according to the attached image, we can know both variables:

The center is in the coordinates (4,4).

The radius is the difference between the "x" center coordinate and "x" B coordinate:

r=7-4=3

Then, we replace the values in the standard form. If we expand both square of the binomial, we get the general form of the equation.

$(x-4)^2+(y-4)^2=3^2\\x^2-8x+16+y^2-8y+16=9\\x^2+y^2-8x-8y+32-9=0\\x^2+y^2-8x-8y+23=0$

Finally, the answer is A. $x^2+y^2-8x-8y+23=0$