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

##### 1 Answer

#### Answer:

Centre Point at:

Radius of: 5

#### Explanation:

The function is a circular function:

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

Where the centre point of the graph is present at the point

and the solution (the number after the = sign) is the radius of the circle squared.

Therefore, from your function:

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

By simplifying this and solving for

**Remember that any real square has two solutions (a positive and negative), hence the #+-sqrt21#**

Therefore, two

One at:

The other at:

**For any function, #y# intercepts where #x# = 0**

Therefore, if we substitute

By simplifying and solving for

**Again, remember that any real square has two solutions (a positive and negative), hence the #+-sqrt16#**

Therefore, two

One at:

The other at:

**To summarise:**

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

Centre Point at:

Radius of: 5

graph{(x+3)^2+(y-2)^2=25 [-21.42, 18.58, -7.32, 12.68]}