Question

How do you graph `(x+3)^2 + (y-2)^2 = 25`?

Answer 1

`x` intercepts at: `x=+-sqrt21-3`
`y` intercepts at: `y= +-4+2`
Centre Point at: `(-3,2)`
Radius of: 5

Explanation:

The function is a circular function:

The general form of a circular function can be expressed as:

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

Where the centre point of the graph is present at the point `(h,k)`
and the solution (the number after the = sign) is the radius of the circle squared.

Therefore, from your function:

`(x+3)^2+(y-2)^2=25`

We can determine that:

The centre point of the function is present at the point `(-3,2)` and the radius is `sqrt25=5`

For any function, `x` incepts where `y` = 0

Therefore, by substituting `y=0` we get:

`(x+3)^2+(-2)^2=25`

By simplifying this and solving for `x` we get:

`(x+3)^2+4=25`
`(x+3)^2=25-4`
`(x+3)^2=21`
`(x+3)=+-sqrt21`
Remember that any real square has two solutions (a positive and negative), hence the `+-sqrt21`
`x=+-sqrt21-3`
Therefore, two `x` intercepts are present:
One at: `x=-sqrt21-3`
The other at: `x=+sqrt21-3`

For any function, `y` intercepts where `x` = 0

Therefore, if we substitute `x` = 0 into your equation, we get:

`(0+3)^2+(y-2)^2=25`

By simplifying and solving for `y` we get:

`9+(y-2)^2=25`
`(y-2)^2=16`
Again, remember that any real square has two solutions (a positive and negative), hence the `+-sqrt16`
`(y-2)=+-sqrt16`
`(y-2)=+-4`
`y=+-4+2`
Therefore, two `y` intercepts are present:
One at: `y=-2` calculated as: `[-4+2]`
The other at: `y=6` calculated as: `[+4+2]`

To summarise:

If we plot all of our points on the graph, we get:

Centre Point at: `(-3,2)`

Radius of: 5

`x` Intercepts at: `x= +sqrt21-3` and `x=-sqrt21-3`

`y` Intercepts at: `y=-2` and `y=6`
graph{(x+3)^2+(y-2)^2=25 [-21.42, 18.58, -7.32, 12.68]}