To answer this question, we need to remember the equation for a circle with a center at point (h, k) is given by:

We have that the center of this circle is located at point (0, 4). Then, we have:
• h = 0
,
• k = 4
And we also know that the point (-2, -1) lies on this circle. Then, we have:
• x = -2
,
• y = -1
Therefore, we can substitute these values into the previous equation to find the radius of this circle as follows:

Now, we finally have that the length of the radius of this circle is (squared root of 29 units) (third option):
![r^2=29\Rightarrow\sqrt[]{r^2}=\sqrt[]{29}\Rightarrow r=\sqrt[]{29}](https://img.qammunity.org/2023/formulas/mathematics/college/2f7kzp68n8kjnq71v8k5b159m4mgvj4oua.png)
We can represent graphically the equation of the circle as follows:
![(x-0)^2+(y-4)^2=(\sqrt[]{29})^2=(x-0)^2+(y-4)^2=29](https://img.qammunity.org/2023/formulas/mathematics/college/6mxp7zxptqtrusf8ipvlf5qrxvfzpevmom.png)