Question

Find the equation of a circle which has D (4, 1) and F (-2, -7) as its diameter.

Leave your answer in the form x² + y² + 2gx + 2fy + c = 0

Answer 1

Check the picture below.

so the center of it as at the midpoint of DF and the distance from that center to either D or F is its radius.

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ D(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\qquad F(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-7}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -2 +4}{2}~~~ ,~~~ \cfrac{ -7 +1}{2} \right) \implies \left(\cfrac{ 2 }{2}~~~ ,~~~ \cfrac{ -6 }{2} \right)\implies \stackrel{center}{(1~~,~~-3)}`

now from there to say hmmm point D

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{center}{(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})}\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{1})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ d=√((~~4 - 1~~)^2 + (~~1 - (-3)~~)^2) \implies d=√((4 -1)^2 + (1 +3)^2) \\\\\\ d=√(( 3 )^2 + ( 4 )^2) \implies d=√( 9 + 16 ) \implies d=√( 25 )\implies \stackrel{radius}{d=5}`

`~\dotfill\\\\ \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{1}{h}~~,~~\underset{-3}{k})}\qquad \stackrel{radius}{\underset{5}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - 1 ~~ )^2 ~~ + ~~ ( ~~ y-(-3) ~~ )^2~~ = ~~5^2\implies (x-1)^2 + (y+3)^2 = 25 \\\\\\ (x^2-2x+1) + (y^2 +6y+9)-25=0\implies \boxed{x^2+y^2-2x+6y-15=0}`