How do you write the equation of the circle with endpoints of its diameter at (-4,7) and (8,-9)?

2 Answers
Oct 6, 2017

(x-2)^2+(y+1)^2=100

Explanation:

"the standard form of the equation of a circle is"

color(red)(bar(ul(|color(white)(2/2)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(2/2)|)))

"where "(a,b)" are the coordinates of the centre and r is"
"the radius"

"to find the centre we require the "color(blue)"midpoint "" of the"
"2 given points"

"centre "=[1/2(-4+8),1/2(7-9)]

color(white)(centre)=(2,-1)

"the radius is the distance from the centre to either of "
"the 2 given points"

"calculate the radius using the "color(blue)"distance formula"

•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

"let "(x_1,y_1)=(2,-1)" and "(x_2,y_2)=(-4,7)

r=sqrt((-4-2)^2+(7+1)^2)=sqrt(36+64)=10

rArr(x-2)^2+(y-(-1))^2=10^2

rArr(x-2)^2+(y+1)^2=100larr" equation of circle"

Oct 6, 2017

(x-2)^2+(x+1)^2=100

Explanation:

If endpoint of the diameter are (-4,7) and (8,-9)
then the circle has

  • a center at (color(red)(x_c),color(blue)(y_c))=((-4+8)/2,(7+(-9))/2)=(color(red)2,color(blue)(-1))
    and
  • a diameter of color(green)d=sqrt((-4-(+8))^2+(7-(-9))^2)=sqrt(12^2+16^2)=sqrt(144+256)=sqrt(400)=20
    rarr a radius of color(magenta)r=color(green)d/2=color(magenta)10

The general equation of a circle with center (color(red)(x_c),color(blue)(y_c)) and radius color(magenta)r is
color(white)("XXX")(x-color(red)(x_c))^2+(y-color(blue)(y_c))^2=color(magenta)r^2

In this case the equation will be:
color(white)("XXX")(x-2)^2+(x+1)^2=100