Question

Find the equation of the circle in general forn if the diameter is 12 and the center is 8,1? Thabk you for the answer

Answer 1

`(x-8)^2+(y-1)^2=6^2`

Explanation:

If the diameter is `12` then the radius is `12/2=6`

The general equation of a circle with center `(a,b)` and radius `r` is
`color(white)("XXX")(x-a)^2+(y-b)^2=r^2`

Answer 2

`(x-8)^2+(y-1)^2=6^2`

Explanation:

The equation of a circle is basically using Pythagoras `a^2+b^2=c^2`

It is just that we are using the x-axis value as say `a`, the y-axis value as say `b` and the radius of the circle as `c`

If the centre is at the origin we have `x^2+y^2=r^2`

This works very well when centred at at the origin but what if it is elsewhere?

We 'hypothetically' take it back to the origin

So if the centre is at 8 on the x-axis we write `(x-8)`
So if the centre is at 1 on the y-axis we write `(y-1)`

The length of the radius is unchanged giving:

`(x-8)^2+(y-1)^2=r^2=(12/2)^2`