If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle?

Accepted Solution

Equation of the circle is [tex](x+6)^{2}+(y+10)^{2}=20[/tex].Solution:The endpoints of the diameter of a circle are (–8, –6) and (–4, –14).Center of the circle = Mid point of the diameterMid point formula:[tex]$P(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)[/tex]Here, [tex]x_1=-8, y_1=-6, x_2=-4, y_2=-14[/tex][tex]$P(x, y) =\left(\frac{-8-4}{2}, \frac{-6-14}{2}\right)[/tex][tex]$P(x, y) =\left(\frac{-12}{2}, \frac{-20}{2}\right)[/tex][tex]$P(x, y) =(-6, -10)[/tex]Center of the circle = (–6, –10)Radius is the distance between center and any endpoint of the diameter.To calculate the radius using distance formula.[tex]r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}[/tex]Here, [tex]x_1=-6, y_1=-10, x_2=-8, y_2=-6[/tex][tex]r=\sqrt{\left(-8-(-6)\right)^{2}+\left(-6-(-10)}\right)^{2}}[/tex][tex]r=\sqrt{(-8+6)^{2}+(-6+10)^{2}}[/tex][tex]r=\sqrt{(-2)^{2}+(4)^{2}}[/tex][tex]r=\sqrt{20}[/tex] unitsThe standard form of the equation of a circle is[tex](x-a)^{2}+(y-b)^{2}=r^{2}[/tex], where (a, b) are center and r is the radius.Here, center = (–6, –10) and [tex]r=\sqrt{20}[/tex] [tex](x-(-6))^{2}+(y-(-10))^{2}={(\sqrt{20})} ^{2}[/tex][tex](x+6)^{2}+(y+10)^{2}=20[/tex]Equation of the circle is [tex](x+6)^{2}+(y+10)^{2}=20[/tex].