Precalculus Exam 1 Review: Circles, Complex Numbers, and Coordinate Geometry

Q10. Find the center (h, k) and radius r of the circle. Graph the circle.

Given the equation: $x^2 + y^2 - 2x - 8y - 8 = 0$

Background

Topic: Circles in the Coordinate Plane

This question tests your ability to rewrite the equation of a circle in standard form by completing the square, and then identify the center and radius. You are also asked to match the equation to its correct graph.

Key Terms and Formulas

• Standard form of a circle: $(x - h)^2 + (y - k)^2 = r^2$

• Center: $(h, k)$

• Radius: $r$

• Completing the square: A method to rewrite quadratic expressions in the form $(x - h)^2$ or $(y - k)^2$.

Step-by-Step Guidance

1. Start by grouping the $x$ terms and $y$ terms together: $x^2 - 2x + y^2 - 8y = 8$

2. To complete the square for $x^2 - 2x$, take half of the coefficient of $x$ (which is $-2$), square it, and add it to both sides. Do the same for $y^2 - 8y$.

3. Write the completed squares as $(x - h)^2$ and $(y - k)^2$ on the left side, and simplify the right side to find $r^2$.

4. Identify the values of $h$, $k$, and $r$ from your standard form equation.

5. Now, look at the provided graphs and match the center and radius you found to the correct graph.

Try solving on your own before revealing the answer!

Final Answer:

The standard form is $(x - 1)^2 + (y - 4)^2 = 25$.

So, the center is $(h, k) = (1, 4)$ and the radius is $r = 5$.

The correct graph is the one with center at $(1, 4)$ and radius $5$.

We completed the square for both $x$ and $y$ terms, rewrote the equation in standard form, and matched it to the correct graph.