Question

Find the equation of the circle with A(2,-3)and B(-3,5)as endpoints of a diameter?

Answer 1

To find the equation of a circle, we need to find the radius as well as the center.

Since we have the endpoints of the diameter, we can use the midpoint formula to obtain the midpoint, which also happens to be the center of the circle.
Finding the midpoint:
`M=((2+(-3))/2,(-3+5)/2)=(-1/2,1)`
So the center of the circle is `(-1/2,1)`

Finding the radius:
Since we have the endpoints of the diameter, we can apply the distance formula to find the length of the diameter. Then, we divide the length of the diameter by 2 to obtain the radius. Alternatively, we can use the the co-ordinates of the center and one of the endpoints to find the length of the radius (I'll leave this to you - the answers will be the same).

`AB = sqrt((2-(-3))^2 + (-3-5)^2)`
`:. AB=sqrt(89)`
`radius=sqrt(89)/2`

The general equation of a circle is given by:
`(x-a)^2+(y-b)^2=r^2`
So we have,
`(x-(-1/2))^2+(y-1)^2=(sqrt(89)/2)`
Therefore, the equation of the circle is `(x+1/2)^2+(y-1)^2=89/4`

Answer 2

`x^2+y^2+x-2y-21=0`

Explanation:

The equation of the circle with `A(x_1,y_1) and B(x_2,y_2)` as

endpoints of a diameter is

`color(red)((x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0)`.

We have , `A(2,-3) and B(-3,5).`

`:.` The required equn.of the circle is,

`(x-2)(x+3)+(y+3)(y-5)=0`.

`=>x^2+3x-2x-6+y^2-5y+3y-15=0`

`=>x^2+y^2+x-2y-21=0`

Answer 3

`(x+1/2)^2+(y-1)^2=89/4`

Very full explanation given

Explanation:

There are two things to solve hear.

1: what is the radius (we will need that)

2: where is the centre of the circle.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the centre point")`

This will be the mean values of x's and the mean of the y's

Mean value of `x`: we go from -3 to 2 which is a distance of 5. Halve of this distance is `5/2` so we have:

`x_("mean") = -3+5/2 = -1/2`

Mean value of `y`: we go from -3 to 5 which is 8. Half of 8 is 4 so we have: `-3+4=+1`

`color(red)("Centre point "->(x,y)=( -1/2,+1))`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the radius")`

We use Pythagoras to determine the distance between the points

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

`D=sqrt([2-(-3)]^2+[-3-5]^2)`

`D=sqrt(25+64 )=sqrt(89)` Note that 89 is a prime number

`color(red)("So radius "->r=D/2=sqrt(89)/2~~4.7169905..." Approximately")`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the equation of the circle")`

This is not what is really happening but what follows will help you remember the equation.

If the centre is at `(x,y)=(-1/2,1)` then if we move this point back to the origin (crossing of the axis) we have:

`(x+1/2) and (y-1)`

To make this into the equation of a circle we use Pythagoras (again) giving:

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

But we know that `r=sqrt(89)/2" so " r^2= 89/4` giving:

`(x+1/2)^2+(y-1)^2=89/4`