Question

How do we find the standard equation of the circle with endpoints of a diameter `(1/2, 4)` `(3/2, -1)`?

Answer 1

`(x-1)^2+(y-3/2)^2=2.55^2`

Explanation:

The standard equation of a circle takes the form:

`(x-a)^2+(y-b)^2=c^2` where `c` is the radius

We need to find the radius, `r`, and the location of the centre, the point `(a,b)`

The length of the diameter is given by:

`d = 2r = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)`

`d = sqrt((3/2-1/2)^2 + (-1-4)^2)`
`= sqrt(1^2+(-5)^2) = sqrt(1+25) = 5.1` units

The radius will be half the diameter, `2.55` units

We can then use the midpoint formula to find the coordinates of the centre. The midpoint formula simply averages the values of the endpoints:

`(a,b) = ((x_1+x_2)/2,(y_1+y_2)/2) = ((1/2+3/2)/2,(4-1)/2)=(1,3/2)`

Now we have all the information we need to write the equation for the circle:

`(x-1)^2+(y-3/2)^2=2.55^2`

Answer 2

`(x-1)^2+(y-3/2)^2=13/2`

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 the radius.

`"the centre will be at the midpoint of the diameter"`

`"midpoint "=[1/2(1/2+3/2),1/2(4-1)]"`

`rArr"centre " =(1,3/2)larr(a,b)`

`"the radius is the distance from the centre to either of "`
`"the endpoints"`

`"to calculate r use the "color(blue)"distance formula"`

`"points are " (1,3/2)" and " (3/2,-1)`

`r=sqrt((3/2-1)^2+(-1-3/2)^2)`

`color(white)(r)=sqrt(1/4+25/4)=sqrt(13/2)`

`rArr(x-1)^2+(y-3/2)^2=(sqrt(13/2))^2`

`rArr(x-1)^2+(y-3/2)^2=13/2" is the equation"`