Find the circumference of the given circle
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[tex]\huge\boxed{2\pi\sqrt{41}}[/tex]
Hey! Start by finding the radius. We will assume that point [tex]P[/tex] is the center point of the circle, so the radius is the distance between points [tex]P[/tex] and [tex]Q[/tex].
Let's use the distance formula, substituting in the known values:
[tex]\begin{aligned}r&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\&=\sqrt{(-2-3)^2+(2-(-2))^2}\end{aligned}[/tex]
Simplify:
[tex]\begin{aligned}r&=\sqrt{(-2-3)^2+(2-(-2))^2}\\&=\sqrt{(-5)^2+4^2}\\&=\sqrt{25+16}\\&=\sqrt{41}\end{aligned}[/tex]
Now, we'll use the formula for the circumference of a circle, substituting in the known value:
[tex]\begin{aligned}C&=2\pi r\\&=\boxed{2\pi\sqrt{41}}\\&\approx40.232\end{aligned}[/tex]
Answer:
40.21
Step-by-step explanation:
The circumference of a circle is 2πr; where are is the radius. The radius is not given but two points
(x1,y1) = (-2,2)
(x2,y2) = (3,-2)
With the two points you can use the distance formula which will be equal to the radius
r = [tex]\sqrt{(y2-y1)^{2} + (x2-x1)^{2} }[/tex]
r = [tex]\sqrt{(-2-2)^{2} + (3+2)^{2} }[/tex]
r = [tex]\sqrt{(-4)^{2} + (5)^{2} }[/tex]
r = [tex]\sqrt{41 }[/tex] = 6.40
circumference = 2πr = 2π(6.40) = 40.21