Question

Identify the center and radius of the circle (x + 2)^2 + (y – 3)^2 = 9
Check all that apply

Center: (2, -3) r = 3

Center: (2, -3) r = 9

Center: (-2, 3) r = 3

Center: (-3, 2) r = 9


What is the radius of a circle with the equation x^2 + y^2 + 2x + 4y – 9 = 0
Round your answer to the nearest thousandth. _____________


Which of the following is a degenerate circle?
Check all that apply

x + y = 7

x^2 + y^2 = -2

x^2 + y^2 = 5

(x – 5)^2 + (y – 3)^2 = 0

Answer 1

Answer on Question #57172 – Math – Analytic Geometry

Question

1. Identify the center and radius of the circle $(x + 2)^2 + (y - 3)^2 = 9$. Check all that apply:

- Center: $(2, -3)$, $r = 3$

- Center: $(2, -3)$, $r = 9$

- Center: $(-2, 3)$, $r = 3$

- Center: $(-3, 2)$, $r = 9$

Solution

Equation of the circle:

$$(x - a)^2 + (y - b)^2 = r^2,$$

where $O(a; b)$ is the center, $r$ is the radius.

Equation of the circle in this question^

$$(x - (-2))^2 + (y - 3)^2 = 3^2,$$

where $(-2, 3)$ is the center, $r = 3$ is the radius.

**Answer**: center: $(-2, 3)$, radius: $r = 3$.

Question

2. What is the radius of a circle with the equation $x^2 + y^2 + 2x + 4y - 9 = 0$. Round your answer to the nearest thousandth.

Solution

Equation of the circle:

$$(x - a)^2 + (y - b)^2 = r^2$$ $$x^2 + y^2 + 2x + 4y - 9 = 0$$ $$x^2 + 2x + y^2 + 4y = 9$$ $$(x^2 + 2 \cdot 1 \cdot x + 1^2) + (y^2 + 2 \cdot 2 \cdot y + 2^2) = (1^2 + 2^2 + 9)$$ $$(x + 1)^2 + (y + 2)^2 = 14$$ $$(x + 1)^2 + (y + 2)^2 = (\sqrt{14})^2$$ $$r = \sqrt{14} \approx 3.742.$$

**Answer**: $r = 3.742$.

Question

3. Which of the following is a degenerate circle? Check all that apply:

$$x + y = 7$$ $$x^2 + y^2 = -2$$ $$x^2 + y^2 = 5$$ $$(x - 5)^2 + (y - 3)^2 = 0$$

Solution

A degenerate circle is a circle of zero radius.

$x + y = 7$ is a straight line.

$x^2 + y^2 = -2$ is an imaginary circle, because $r^2 = -2 < 0$.

$x^2 + y^2 = 5$ is a circle, $r = \sqrt{5}$.

$(x - 5)^2 + (y - 3)^2 = 0$ is a degenerate circle, because $r = 0$.

**Answer**: $(x - 5)^2 + (y - 3)^2 = 0$ is a degenerate circle.

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