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The points (7,13) and (-3,-11) are at the ends of a diameter. Find an equation for this circle.

 Jun 30, 2015

Best Answer 

 #1
avatar+128079 
+10

The radius = (1/2)√[ (-3-7)^2 + (-11 - 13)^2] = (1/2)√[10^2 + 24^2]  = (1/2)√676  = (1/2)26  = 13

 

And the center is at   [ ( 7 -3) / 2 ,  (13-11) / 2 ]   =  [4/2,  2.2] =  (2, 1)   and the equation is....

 

(x - 2)^2  + (y - 1)^2  = 13^2       ..... or......

 

(x - 2)^2  + (y - 1)^2  = 169

 

Here's a graph........https://www.desmos.com/calculator/rx1yothteb

 

 

 Jul 1, 2015
 #1
avatar+128079 
+10
Best Answer

The radius = (1/2)√[ (-3-7)^2 + (-11 - 13)^2] = (1/2)√[10^2 + 24^2]  = (1/2)√676  = (1/2)26  = 13

 

And the center is at   [ ( 7 -3) / 2 ,  (13-11) / 2 ]   =  [4/2,  2.2] =  (2, 1)   and the equation is....

 

(x - 2)^2  + (y - 1)^2  = 13^2       ..... or......

 

(x - 2)^2  + (y - 1)^2  = 169

 

Here's a graph........https://www.desmos.com/calculator/rx1yothteb

 

 

CPhill Jul 1, 2015

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