Question

There is a circle of radius 6 and its centre is at `(x,y)=(3,7)`. What is the equation for this circle?

Answer 1

`(x-3)^2+(y-7)^2=6^2`

Explanation:

Think of Pythagoras `a^2+b^2=c^2`
Compare the circle `color(white)("d")x^2+y^2=r^2`

The above is with the centre of the circle at the origin.

This bit is worth remembering. If you wish to 'move' the `x` to the right you subtract a value. On the other hand if you wish to move the `x` to the left you add a value. Similarly with the y

Thus changing the centre to `(x,y)=(3,7)` you have 'shifted' the centre 3 to the left of origin and down 7 below the origin.

Thus we have `x-3 and y-7 -> (x-3)^2+(y-7)^2=6^2`

Answer 2

`(x-3)^2+(y-7)^2=36`

Explanation:

The equation of a circle is given by `(x-a)^2+(y-a)^2=r^2`, where:

• `a` is the `x` coordinate of the center.
• `b` is the `y` coordinate of the center.
• `r` is the radius.

Here, `r=6`, `a=3` and `b=7`

So, the equation is `(x-3)^2+(y-7)^2=6^2`

`(x-3)^2+(y-7)^2=36`

Proof (you can press edit answer and look for yourself):

graph{((x-3)^2+(y-7)^2-36)(y-7)(x-3)=0 [-9.29, 15.54, 0, 12.42]}