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# Coordinates

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The points (2,3) and (3,2) lie on a circle whose center is on the x-axis. What is the radius of the circle?

Apr 14, 2022

#1
+1384
+1

Because of the shape of a circle, both the points have to be the same distance from the center.

Using the Pythagorean theorem, we can write this as an equation: $$\sqrt{(2-x)^2 + (3-y)^2} = \sqrt{(3-x)^2+(2-y)^2}$$

Substituting 0 for y, we get: $$(2-x)^2 +9 = (3-x)^2+4$$

Solving, we find $$x = 0$$, meaning the coordinates of the center are: $$(0, 0)$$

Using the Pythagorean Theorem, we find that the radius of the circle is $$\color{brown}\boxed{\sqrt{13}}$$

Apr 14, 2022

#1
+1384
+1

Because of the shape of a circle, both the points have to be the same distance from the center.

Using the Pythagorean theorem, we can write this as an equation: $$\sqrt{(2-x)^2 + (3-y)^2} = \sqrt{(3-x)^2+(2-y)^2}$$

Substituting 0 for y, we get: $$(2-x)^2 +9 = (3-x)^2+4$$

Solving, we find $$x = 0$$, meaning the coordinates of the center are: $$(0, 0)$$

Using the Pythagorean Theorem, we find that the radius of the circle is $$\color{brown}\boxed{\sqrt{13}}$$

BuilderBoi Apr 14, 2022
#2
+122
-3

Good BØ¥¥¥¥

Kakashi  Apr 14, 2022