MATH SOLVE

10 months ago

Q:
# Write the equation of the circle with center (−1, −3) and (−7, −5) a point on the circle.

Accepted Solution

A:

so, we know the center is at -1, -3, hmmm what's the radius anyway?

well, the radius will be the distance from the center to any point on the circle, it just so happen that we know -7, -5 is on it, thus

[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -1 &,& -3~) % (c,d) &&(~ -7 &,& -5~) \end{array} \\\\\\ d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ r=\sqrt{[-7-(-1)]^2+[-5-(-3)]^2}\implies r=\sqrt{(-7+1)^2+(-5+3)^2} \\\\\\ r=\sqrt{36+4}\implies r=\sqrt{40}\\\\ -------------------------------[/tex]

[tex]\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-1}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{\sqrt{40}}{ r} \\\\\\\ [x-(-1)]^2+[y-(-3)]^2=(\sqrt{40})^2\implies (x+1)^2+(y+3)^2=40[/tex]

well, the radius will be the distance from the center to any point on the circle, it just so happen that we know -7, -5 is on it, thus

[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -1 &,& -3~) % (c,d) &&(~ -7 &,& -5~) \end{array} \\\\\\ d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ r=\sqrt{[-7-(-1)]^2+[-5-(-3)]^2}\implies r=\sqrt{(-7+1)^2+(-5+3)^2} \\\\\\ r=\sqrt{36+4}\implies r=\sqrt{40}\\\\ -------------------------------[/tex]

[tex]\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-1}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{\sqrt{40}}{ r} \\\\\\\ [x-(-1)]^2+[y-(-3)]^2=(\sqrt{40})^2\implies (x+1)^2+(y+3)^2=40[/tex]