The points (3,-2) and (-2,3) lie on a circle whose center is on the x-axis. What is the radius of the circle?
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+9466 Let the center of the circle be (h, 0)
distance between (h, 0) and (3, -2) = distance between (h, 0) and (-2, 3)
\(\begin{array}{r} \sqrt{(-2-0)^2+(3-h)^2}&\ =&\ \sqrt{(-2-h)^2+(3-0)^2}\\~\\ (-2-0)^2+(3-h)^2&=&(-2-h)^2+(3-0)^2\\~\\ (-2-0)^2-(3-0)^2&=&(-2-h)^2-(3-h)^2 \end{array}\)
Notice here that if h = 0 both sides of the equation are identical.
If you want to, you can expand both sides to also find that h = 0.
So the center of the circle is (0, 0)
And the radius = \(\sqrt{(-2-0)^2+(3-0)^2}\ =\ \sqrt{4+9}\ =\ \sqrt{13}\)
Here's a graph: https://www.desmos.com/calculator/najnuubd4e
+9466 Let the center of the circle be (h, 0)
distance between (h, 0) and (3, -2) = distance between (h, 0) and (-2, 3)
\(\begin{array}{r} \sqrt{(-2-0)^2+(3-h)^2}&\ =&\ \sqrt{(-2-h)^2+(3-0)^2}\\~\\ (-2-0)^2+(3-h)^2&=&(-2-h)^2+(3-0)^2\\~\\ (-2-0)^2-(3-0)^2&=&(-2-h)^2-(3-h)^2 \end{array}\)
Notice here that if h = 0 both sides of the equation are identical.
If you want to, you can expand both sides to also find that h = 0.
So the center of the circle is (0, 0)
And the radius = \(\sqrt{(-2-0)^2+(3-0)^2}\ =\ \sqrt{4+9}\ =\ \sqrt{13}\)
Here's a graph: https://www.desmos.com/calculator/najnuubd4e
