if the given is (10, -16) and the radius’s is $$  $$$$ \sqrt3 $$

Given the center $(10, -16)$ and the radius $ \sqrt{3} $, we want to find the standard form of the equation of an ellipse. Since the radius is equal in all directions, the equation will represent a circle.

The general equation for a circle with center $(h, k)$ and radius $r$ is:

\[
(x - h)^2 + (y - k)^2 = r^2
\]

### Given:
- Center: $(h, k) = (10, -16)$
- Radius: $r = \sqrt{3}$

Substitute these values into the equation:

\[
(x - 10)^2 + (y + 16)^2 = (\sqrt{3})^2
\]

Simplifying:

\[
(x - 10)^2 + (y + 16)^2 = 3
\]

This is the standard form of the equation of a circle with center $(10, -16)$ and radius $\sqrt{3}$.

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Would you like more details or have any questions?

Here are some related questions you might be interested in:

1. How would the equation change if the radius were different for the x and y axes?
2. What is the distance formula and how does it relate to the equation of a circle?
3. How do you find the circumference of a circle given the radius?
4. How would you convert this equation to the general form of a circle?
5. How do you determine if a point lies on the circle?
6. What is the area of this circle?
7. How does the center affect the position of the circle on the coordinate plane?
8. How do you derive the equation of a circle from the distance formula?

**Tip:** When working with square roots in equations, remember that squaring them simplifies the expression, making the equation easier to handle.

Learn how to derive the standard form of the equation of a circle with center (10, -16) and radius √3. This problem explores the concept of circle equations in coordinate geometry and includes a step-by-step solution. Suitable for Grades 9-12.

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