Question

Determine the equation of a circle having a diameter with endpoints at (12,-14) and (2,4)

Answer 1

$The \ equation \ of \ a \ circle \ with \ centre \ (a,b) \ and \ radius \ "r" \ is : \\ \\(x-a)^2+(y-b)^2=r^2$

$(12,-14) , \ \ \ (2,4)\\Since \ the \ center \ of \ the \ circle \ is \ the \ midpoint \ of \ the \ line \ segment \\ connecting \ two \ endpoints \ of \ a \ diameter \\\\Midpoint \ Formula \\\\(a,b)=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})=(\frac{12+2}{2},\frac{-14+4}{2})=(\frac{14}{2},\frac{-10}{2})=(7,-5)$

$The \ radius \ is \ the \ distance \ from \ the \ center \ to \ some \ point \ on \ the \ circle.\\ The \ distance \ from \ (7,-5) \ to \ (12, -14) \ is: \\ \\ r= \sqrt{(x_{2}-x_{1})^2 +(y_{2}-y_{1})^2}\\\\r= \sqrt{(12-7)^2 +(-14+5)^2}=\sqrt{5^2+(-9)^2}=\sqrt{25+81}=\sqrt{106}\\\\(x-7)^2+(y-(-5))^2= (\sqrt{106})^2 \\ \\(x-7)^2+(y+5)^2=106$