Question

which point is not on the circle with center (3,2) and radius 5?
(8, -2)
(-2, 0)
(4, 4)
(8, -2)

Explanation:

Step1: Recall circle equation

The standard equation of a circle with center $(h,k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$. Substitute $h=3$, $k=2$, $r=5$:
$$(x-3)^2+(y-2)^2=25$$

Step2: Test point (8,-2)

Substitute $x=8$, $y=-2$ into the equation:

$$(8-3)^2+(-2-2)^2=5^2+(-4)^2=25+16=41 eq25$$

Step3: Test point (-2,2)

Substitute $x=-2$, $y=2$ into the equation:
$$(-2-3)^2+(2-2)^2=(-5)^2+0^2=25+0=25$$

Step4: Test point (3,6)

Substitute $x=3$, $y=6$ into the equation:

$$(3-3)^2+(6-2)^2=0^2+4^2=0+16=16 eq25$$

Step5: Test point (6,-2)

Substitute $x=6$, $y=-2$ into the equation:
$$(6-3)^2+(-2-2)^2=3^2+(-4)^2=9+16=25$$

Answer:

A. (8, -2), C. (3, 6)