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Find the standard form for the equation of a circle (x-h)^2+(y-k)^2=r^2 with a diameter that has endpoints (-5,0) and (8,-6).

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2 Answers

Dear Jamey,
     Robert did a fine job of getting the answer; however, I would like you to take note of something.  You  may find
it interesting to graph the two end points (Remember that endpoints are on the circumference of the circle.) on some
graph paper.  The segment that can be drawn from the endpoints creates the diameter of the circle.  We don't know
the midpoint of this segment, which is also the midpoint of the circle.The next thing to do is to construct a right
triangle.  If draw a perpendicular segment(It intercepts 90 ° with the triangle's base) from (8,6) to the point (8,0),
you have one side of a right triangle.  If you count the units up, you get 6 units for the distance. Now you can also
count the units from (8,0) to (-5, 0).  That comes to 13 units.  Then you can find the diameter of the circle by finding
the distance of the hypotenuse of the right triangle.  
Use Pythagorean Theorem, which gives you 62 +132 = C2.     36 +169=C2     205=C2=r2
Now find C by taking the square root of both sides.    C=Diameter of circle= D=√(205) , so   r=1/2 (D)
r=  D/2   =√(205) /2       X midpoint of diameter = (-5+8)/2        Y midpoint of diameter= (0+6)/2
midpoint coordinates (3/2, 6/2)  = (1.5,3)
(h,K)= center of circle= midpoint of diameter= (1.5,3)      C2=r2= (√(205) /2) 2 = 205/4

Now here is the equation  (X-h)2 + (Y-K)2= C2=r2.  Use substitution to rewrite it.
                                    (X-1.5)2 + (Y-3)2= C2= 205/4    Answer
Use midpoint theorem to find the center (h, k):
h = (-5+8)/2 = 3/2, and k = (0-6)/2 = -3
Find the diameter,
d = sqrt(13^2+6^2) = sqrt(205)
r = d/2 = (1/2)sqrt(205)
Answer: (x - 3/2)^2 + (y+3)^2 = 205/4


   Please check your "Y" value in your midpoint.  I believe that you subtracted, instead of of adding.