SOLUTION: Find equation of the circle whose diameter has endpoints (1,-6) and (-9,0)

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Question 1039382: Find equation of the circle whose diameter has endpoints (1,-6) and (-9,0)
Found 3 solutions by josgarithmetic, ikleyn, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Without refinement but using standard form, Midpoint formula, AND the Distance Formula,




Most people would be more comfortable finding each part separately, before putting the finished pieces into the standard form equation.

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find equation of the circle whose diameter has endpoints (1,-6) and (-9,0)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The center is in the point  x%5B0%5D = %28%28-9%29%2B1%29%2F2 = -4,  y%5B0%5D = %28%28-6%29+%2B+0%29%2F2 = -3.


The diameter of the circle is  d = sqrt%28%28%28-9%29-1%29%5E2+%2B+%280-%28-6%29%29%5E2%29 = sqrt%28%28-10%29%5E2+%2B+6%5E2%29 = sqrt%28100+%2B+36%29 = sqrt%28136%29.


Hence, the radius of the circle is   r = sqrt%28136%29%2F2 = %282%2Asqrt%2834%29%29%2F2 = sqrt%2834%29.


Finally, the equation of the circle is 


%28x-%28-4%29%29%5E2+%2B+%28y-%28-3%29%29%5E2 = r%5E2,  or


%28x%2B4%29%5E2+%2B+%28y%2B3%29%5E2 = 34.


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Find equation of the circle whose diameter has endpoints (1,-6) and (-9,0)
1st: Find the CENTER, which is the midpoint of the diameter
() =====> (matrix%281%2C3%2C+%281+%2B+-+9%29%2F2%2C+%22%2C%22%2C+%28-+6+%2B+0%29%2F2%29) =====> matrix%281%2C5%2C+%22%28%22%2C+-+4%2C+%22%2C%22%2C+-+3%2C+%22%29%22%29 =====> matrix%281%2C5%2C+%22%28%22%2C+h%2C+%22%2C%22%2C+k%2C+%22%29%22%29
2nd: Use the CENTER, or (h, k) and one of the 2 points to determine r%5E2, or the %28radius%29%5E2
%28x+-+h%29%5E2+%2B+%28y+-+k%29%5E2+=+r%5E2
%28-+9+-+-+4%29%5E2+%2B+%280+-+-+3%29%5E2+=+r%5E2
%28-+9+%2B+4%29%5E2+%2B+%283%29%5E2+=+r%5E2
%28-+5%29%5E2+%2B+3%5E2+=+r%5E2
25+%2B+9+=+r%5E2____34+=+r%5E2
3rd: Use the CENTER, or (h, k) and r%5E2 to form the equation of the circle
%28x+-+h%29%5E2+%2B+%28y+-+k%29%5E2+=+r%5E2
%28x+-+-+4%29%5E2+%2B+%28y+-+-+3%29%5E2+=+34 ------- Substituting (- 4, - 3) for (h, k), and 34 for r%5E2