Question

chapter 7:7.1-7.2
due: tue jun 23, 2026 11:59pm

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chapter 7:7.1-7.2
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question 1

find the center and radius of the circle whose equation is (x^2 + y^2 + 6y - 18 = 0).

the center of the circle is ( , ).
the radius of the circle is .

question help: video

Explanation:

Step 1: Group the terms

Group the \(x\) and \(y\) terms of the equation.

$$x^2 + (y^2 + 6y) = 18$$

Step 2: Complete the square

Add the square of half the \(y\)-coefficient to both sides.

$$x^2 + (y^2 + 6y + 9) = 18 + 9$$

Step 3: Write in standard form

Simplify the equation into the standard circle form.

$$(x - 0)^2 + (y + 3)^2 = 27$$

Step 4: Identify center and radius

Extract the center \((h, k)\) and radius \(r = \sqrt{27}\).

$$(h, k) = (0, -3), \quad r = 3\sqrt{3}$$

Answer:

The center of the circle is \((0, -3)\).
The radius of the circle is \(3\sqrt{3}\).