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 \`


`(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)=((x_(1)+x_(2))/(2),(y_(1)+y_(2))/(2))=((12+2)/(2),(-14+4)/(2))=((14)/(2),(-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= √((12-7)^2 +(-14+5)^2)=√(5^2+(-9)^2)=√(25+81)=√(106)\\\\(x-7)^2+(y-(-5))^2= (√(106))^2 \\ \\(x-7)^2+(y+5)^2=106`