If an equation has a squared term, and you are asked to solve for the radius and center, it usually suggests that the equation represents a circle. The standard form of the equation of a circle is:

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

Where:

• $(h, k)$ is the center of the circle.
• $r$ is the radius of the circle.

Steps to Solve for Radius and Center:

1. Identify the squared terms: Look for terms like $(x - h)^2$ and $(y - k)^2$.
2. Rewrite the equation: If the equation isn't already in the standard form, you may need to complete the square.
3. Find the center $(h, k)$: The center is given by the values $h$ and $k$ from the squared terms.
4. Calculate the radius: The radius $r$ is the square root of the constant on the right side of the equation.

Example:

Given the equation: $x^2 + y^2 - 4x + 6y - 12 = 0$

1. Rewrite the equation by completing the square:

   • Group $x$ and $y$ terms: $(x^2 - 4x) + (y^2 + 6y) = 12$
   • Complete the square:
     • $x^2 - 4x + 4 = (x - 2)^2$
     • $y^2 + 6y + 9 = (y + 3)^2$
   • Update the equation: $(x - 2)^2 + (y + 3)^2 = 25$

2. Find the center:

   • The center $(h, k)$ is $(2, -3)$.

3. Calculate the radius:

   • The radius $r = \sqrt{25} = 5$.

Final Answer:

• Center: $(2, -3)$
• Radius: $5$

Would you like more details or have any questions?

Related Questions:

1. How do you complete the square for different quadratic expressions?
2. What are the steps to derive the equation of a circle given a center and a point on the circle?
3. How do you find the equation of a circle if you are given the diameter?
4. What happens if the circle's equation is not centered at the origin?
5. How do you solve for the center and radius if the equation is in general form?

Tip:

Always check if the equation can be factored or if completing the square is necessary to convert it into the standard circle form.