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

A
`x^2+y^2-8x-8y+23=0`

Explanation:

From the diagram you can see that the center of the circle is placed at point O(4,4). The radius of the circle is

`r=AO=√((4-4)^2+(7-4)^2)=√(0^2+3^2)=√(9)=3.`

Thus, the equation of the circle is

`(x-4)^2+(y-4)^2=3^2.`

Rewrite it

`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.`