Question

Select the correct answer. What is the general form of the equation of a circle with its center at (-2, 1) and passing through (-4, 1)?

A. x2 + y2 − 4x + 2y + 1 = 0
B. x2 + y2 + 4x − 2y + 1 = 0
C. x2 + y2 + 4x − 2y + 9 = 0
D. x2 − y2 + 2x + y + 1 = 0

Answer 1

the distance from the center of the circle to a point on the circle is simply its radius, thus

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{1})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ r=√((~~-4 - (-2)~~)^2 + (~~1 - 1~~)^2) \implies r=√((-4 +2)^2 + (1 -1)^2) \\\\\\ r=√(( -2 )^2 + ( 0 )^2) \implies r=√( 4 + 0 ) \implies r=√( 4 )\implies \stackrel{ radius }{r=2} \\\\[-0.35em] ~\dotfill`

`\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{-2}{h}~~,~~\underset{1}{k})}\qquad \stackrel{radius}{\underset{2}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - (-2) ~~ )^2 ~~ + ~~ ( ~~ y-1 ~~ )^2~~ = ~~2^2\implies (x+2)^2 + (y-1)^2 = 4 \\\\\\ (x^2+4x+4) + (y^2 - 2y + 1) = 4\implies \boxed{x^2+4x +y^2 -2y + 1 = 0}`