Question

what is the general form of the equation for the given circle? x2 y2 − 8x − 8y 23 = 0 x2 y2 − 8x − 8y 32 = 0 x2 y2 − 4x − 4y 23 = 0 x2 y2 4x 4y 9 = 0

Answer 1

Answer: x2 + y2 − 8x − 8y + 23 = 0

Step-by-step explanation: plato/ edmentum

Answer 2

we know that
the standard form of the equation of the circle is

`(x-h) ^(2) +(y-k) ^(2) =r ^(2)`
where
(h,k) is the center
r is the radius

case a)

`x^(2) + y^(2) -8x-8y+23=0 \\ (x-4) ^(2) +(y-4)^(2) = 3^(2)`
Is a circle
with radius r=
`3` units
and center
`(4,4)`

case b)

`x^(2) + y^(2) -8x-8y+32=0 \\ (x-4) ^(2) +(y-4) ^(2) =0`
Is not a circle

case c)

`x^(2) + y^(2) -4x-y+23=0 \\ (x+2) ^(2) +(y-2) ^(2) =-15`
Is not a circle

case d)

`x^(2) + y^(2) +4x+4y+9=0 \\ (x+2) ^(2) +(y+2) ^(2) =-1`
Is not a circle

therefore

the answer is

`x^(2) + y^(2) -8x-8y+23=0`