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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?

 

THANK U

 Jul 4, 2019

Best Answer 

 #1
avatar+8759 
+5

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

 Jul 4, 2019
 #1
avatar+8759 
+5
Best Answer

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

hectictar Jul 4, 2019

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