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Write the standard form of each circle
Ends of diameter :(-16,-8) (12,4)

Sagot :

AL2006

The ends of the diameter are  (-16, -8) and (12,4)

The point midway between those points is

        x = (1/2) (-16+12) = (1/2) (-4)  =  -2
      
       y = (1/2) (-8+4) = (1/2) ( -4)  =  -2

So the point (-2, -2) is the center of the circle.


The ends of the diameter are  (-16, -8) and (12,4)

The distance between those points is . . .

         the square root of (12+16)² + (4+8)²  =

                   square root of  (28)² + (12)² =

                   square root of  (784 + 144) =

                   square root of    (928)    

The diameter of the circle is  √928 ,
             so its radius is (1/2)√928  =  √232 .

Now that we know the center and the radius,
we can write the equation of the circle:

               (x + 2)² + (y + 2)²  =  232




         



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