Question

The equation of a circle is `x^2+(y+5)^2=81`. What is the center and radius of the circle?

Answer 1

`C(0,-5)`

`r=9`

Explanation:

The equation of a circle is given by `C_1=(x-a)^2+(y-b)^2=r^2`
where `(a,b)` is the centre and `r` is the radius.

We have `C_1=x^2+(y+5)^2=81`

We can rewrite this as `C_1=(x-0)^2+(y-(-5))^2=9^2`. So the centre of `C_1` is `(0,-5)` and the radius is `9`.

Answer 2

Centre`->(x,y)=(x,-5)`
Radius `->sqrt(81)=9`

Explanation:

Proper word for 'moving' is 'transpose'.

Suppose you had some function of `x` that we call `f(x)`

Then this will form a particular shape depending on what the form of `f(x)` is

If we changed this to `f(x-5)` then it will still be the same shape as `f(x)` but it would have 'shifted' right by 5.

On the other hand `f(x+5)` would have 'shifted' the graph of `f(x)` left by 5.

So for the x-axis:

a plus 'shifts' left
a minus 'shifts' right.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Similarly for y-axis:

a plus 'shifts' down
a minus 'shifts' up