Question

Find the equation of the circle with diameter AB where A and B are the points (-1,2) and (3,3) respectively?

Answer 1

The answer could be find out by knowing some formulas of coordinate geometry.
1) standard eqn.=`(x−a)^2 + (y−b)^2 = r^2` (for this question i,e, radius)
2) midpoint= `(x_1+x_2)/2,( y_1+y_2)/2`
3) distance= `sqrt` `(x_2-x_1)^2`+`(y_2-y_1)^2`

Explanation:

we are given , A(-1,2) & B(3,3)
here `x_1`= -1
`y_1`= 2
`x_2`= 3
`y_2`= 3

Therefore, AB= `sqrt` `(x_2-x_1)^2`+`(y_2-y_1)^2`

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

`sqrt` `(4)^2`+`(1)^2`

`sqrt` `(16)`+`(1)`

`sqrt` `17`.....................(Diameter)

so radius = 1/2 * diameter
= `sqrt` `17`/2

now the radius can also be got by calculating the midpoint
so the next step is:
wkt,midpoint is given by =`(x_1+x_2)/2,( y_1+y_2)/2`

then,we get `(-1+3)/2` , `(2+3)/2`

= `(2/2)` , `(5/2)`

= (1 , 2.5)
Hence we get the values of a and b respectively.
Putting the values we get,
`(x−a)^2 + (y−b)^2 = r^2`

= `(x−1)^2`+ `(y−2.5)^2` = `(sqrt`17`/2)^2`
= `(x−1)^2`+ `(y−2.5)^2` = `(17/4)`
= `(x−1)^2`+ `(y−2.5)^2` = 4.25
= `(x−1)^2`+ `(y−2.5)^2` = 4.25........................(answer)

Answer 2

`x^2+y^2-2x-5y+3=0`

Explanation:

Firstly, we can find the centre of the circle by finding the midpoint of `AB`. Since it the midpoint or the centre `(h,k)` cut AB with equal ratio;

`Centre (h,k) = ((-1+3)/2,(2+3)/2)`

`(h,k)=(1,5/2)`

Then we can find the radius , `r` of the circle by using the equation;

`r^2=(x-h)^2+(y-k)^2`

`r^=sqrt((x-h)^2+(y-k)^2`

Substitute the coordinate `(h,k)=(1,5/2)` and any of `A` or `B` coordinates into equation. In this calculation I choose `B`.

`r^=sqrt((3-1)^2+(3-5/2)^2`

`r=sqrt(17/4)`=`sqrt17/2`

To find `c`, we can use the equation `c=h^2+k^2-r^2` and substitute `(h,k)=(1,5/2)` and `r=sqrt17/2` into it.

`c=(1)^2+(5/2)^2-(sqrt17/2)^2`

`c=3`

Then we know that the equation of circle is known as;

`x^2+y^2-2hx-2ky+c=0`

Substitute only `(h,k)=(1,5/2)` and `c=3` and we get;

`x^2+y^2-2(1)x-2(5/2)y+3=0`

`x^2+y^2-2x-5y+3=0`