Question

Question #5e370

Answer 1

The Standard Cartesian Form for the equation of a circle is:
`(x-h)^2+(y-k)^2=r^2" [1]"`
Where `(x,y)` is any point on the circle, `(h,k)` is the center, and `r` is the radius.

Explanation:

Given that the center is `(0,0)` we substitute this into equation [1] to obtain equation [2]:

`(x-0)^2+(y-0)^2=r^2" [2]"`

Substitute the point `(12-5)` into equation [2] and solve for r:

`(12-0)^2+(-5-0)^2=r^2`

`169 = r^2`

`r = 13`

Substitute 13 for r in equation [2]:

`(x-0)^2+(y-0)^2=13^2" [3]"`

Equation [3] is the equation of the specified circle.

Here is its graph:

graph{(x-0)^2+(y-0)^2=13^2 [-30, 30, -15, 15]}