Question

Find the equation of the circle which passes through the origin and `(4, −8)`? `\ \ \ \` Why do we use midpoint of two points to calculate the equation of circle? The points might make a chord rather than a diameter! Give me some insight?

Answer 1

Yes you are right. For detailed discussion, see below.

Explanation:

As one of the points through which circle passes is `(0,0)`

the equation of circle is of the type `x^2+y^2+gx+fy=0`, whose center is `(-g/2,-f/2)`

Further, as it passes through `(4,-8)`, we have

`4^2+(-8)^2+4g-8f=0`

i.e. `g-2f+20=0` .....................(1)

Here we have one equation with two variables `g` and `f`. Therefore we do not have a unique solution. Presently we have infinite solutions, as any point on perpendicular bisector of `(0,0)` and `(4,-8)` could be a center.

Hence we need one more condition, which could be

(a) that `(0,0)` and `(4,-8)` are ends of a diameter (then we do not need (1)). This is what you are referring to in second sentence of your question.

(b) third point, through which circle passes, which will give us second equation in `g` and `f`

or any other condition leading to unique solution for `g` and `f`.

Answer 2

Please see below.

Explanation:

I did not use the midpoint at all.

Using the equation of a circle:

`x^2-2hx+h^2+y^2-2ky+k^2=r^2`

Because it passes through `(0,0)`, we find the `r^2=h^2+k^2`.

So, `x^2-2hx+y^2-2ky=0`

Using the point `(4,-8)`, we get

`16-8h+64+16k=0`

So, `h=2k+10`

Every circle that contains the points `(0,0)` and `(4,-8)` has equation:

`(x-h)^2+(y-k)^2 = h^2+k^2` where `h=2k+10`