Question 1039382
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Find equation of the circle whose diameter has endpoints (1,-6) and (-9,0)
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The center is in the point  {{{x[0]}}} = {{{((-9)+1)/2}}} = {{{-4}}},  {{{y[0]}}} = {{{((-6) + 0)/2}}} = {{{-3}}}.


The diameter of the circle is  {{{d}}} = {{{sqrt(((-9)-1)^2 + (0-(-6))^2)}}} = {{{sqrt((-10)^2 + 6^2)}}} = {{{sqrt(100 + 36)}}} = {{{sqrt(136)}}}.


Hence, the radius of the circle is   r = {{{sqrt(136)/2}}} = {{{(2*sqrt(34))/2}}} = {{{sqrt(34)}}}.


Finally, the equation of the circle is 


{{{(x-(-4))^2 + (y-(-3))^2}}} = {{{r^2}}},  or


{{{(x+4)^2 + (y+3)^2}}} = {{{34}}}.
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