Answer:
[tex](x-4)^2+(y+3)^2=26[/tex]
[tex](x+2)^2+(y-1)^2=26[/tex]
Step-by-step explanation:
A circle passes through the points (-1, -4) and (3, 2) and has radius √(26).
Given that there are two possible equations for this circle, these two circles intersect at the points (-1, -4) and (3, 2).
Let A = (-1, -4) and B = (3, 2). Therefore, line segment AB is a common chord of the two circles. Let M be the midpoint of AB.
The centers of the circles lie on the perpendicular bisector of AB. Since the radii of the two equations are equal, the distance from point M to each center is equal to half of AB.
Find the midpoint of AB:
[tex]\textsf{Midpoint of $\overline{AB}$} =\left(\dfrac{x_B+x_A}{2},\dfrac{y_B+y_A}{2}\right)\\\\\\\textsf{Midpoint of $\overline{AB}$} =\left(\dfrac{3+(-1)}{2},\dfrac{2+(-4)}{2}\right)\\\\\\\textsf{Midpoint of $\overline{AB}$}=(1,-1)[/tex]
Find the slope of AB:
[tex]\textsf{Slope $(m)$}=\dfrac{2-(-4)}{3-(-1)}=\dfrac{3}{2}[/tex]
The slope of a perpendicular line is the negative reciprocal, so the slope of the line perpendicular to AB is m = -2/3.
To find the equation of the line perpendicular to AB, substitute the slope m = -2/3 and the point (1, -1) into the point-slope formula of a linear equation:
[tex]y-(-1)=-\dfrac{2}{3}(x-1)\\\\\\y+1=-\dfrac{2}{3}x+\dfrac{2}{3}\\\\\\y=-\dfrac{2}{3}x-\dfrac{1}{3}[/tex]
The distance between the two centers is equal to the distance between the two points:
[tex]d=\sqrt{(x_M-x_A)^2+(y_M-y_A)^2}\\\\d=\sqrt{(1-(-1))^2+(-1-(-4))^2}\\\\d=\sqrt{(2)^2+(3)^2}\\\\d=\sqrt{4+9}\\\\d=\sqrt{13}[/tex]
To find the x-coordinates of the centers of the two circles, substitute the midpoint M(1, -1), the distance d = √(13), and y = -(2/3)x - 1/3 into the distance formula:
[tex]\sqrt{13}=\sqrt{(1-x)^2+\left(-1-\left(-\dfrac{2}{3}x-\dfrac{1}{3}\right)\right)^2}[/tex]
Solve for x:
[tex](1-x)^2+\left(-1-\left(-\dfrac{2}{3}x-\dfrac{1}{3}\right)\right)^2=13[/tex]
[tex](1-x)^2+\left(\dfrac{2}{3}x-\dfrac{2}{3}\right)^2=13[/tex]
[tex]x^2-2x+1+\dfrac{4}{9}x^2-\dfrac{8}{9}x+\dfrac{4}{9}=13[/tex]
[tex]9x^2-18x+9+4x^2-8x+4=117[/tex]
[tex]13x^2-26x+13=117[/tex]
[tex]13x^2-26x-104=0[/tex]
[tex]13(x^2-2x-8)=0[/tex]
[tex]x^2-2x-8=0[/tex]
[tex]x^2+2x-4x-8=0[/tex]
[tex]x(x+2)-4(x+2)=0[/tex]
[tex](x-4)(x+2)=0[/tex]
Therefore, the x-coordinates of the two centers are x = 4 and x = -2.
To find the corresponding y-coordinates of the centers, substitute the x-coordinates into the equation of the perpendicular line:
[tex]x=4 \implies y=-\dfrac{2}{3}(4)-\dfrac{1}{3}=-3[/tex]
[tex]x=-2 \implies y=-\dfrac{2}{3}(-2)-\dfrac{1}{3}=1[/tex]
Therefore, the centers of the two circles are (4, -3) and (-2, 1).
[tex]\boxed{\begin{array}{l}\underline{\textsf{General equation of a circle}}\\\\(x-h)^2+(y-k)^2=r^2\\\\\textsf{where:}\\ \phantom{ww}\bullet\;\textsf{$(h, k)$ is the center.}\\\phantom{ww}\bullet\;\textsf{$r$ is the radius.}\end{array}}[/tex]
Finally, substitute the centers and the radius into the standard form of the equation for a circle to create the two equations:
[tex](x-4)^2+(y+3)^2=26[/tex]
[tex](x+2)^2+(y-1)^2=26[/tex]
