Question

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

Answer 1

`(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"`

Answer 2

`(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`