6) Write the standard equation of the circle. -6

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
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**Problem 6) Write the standard equation of the circle.**

**Explanation:**
The diagram presents a circle centered at the origin (0, 0) on a coordinate plane. The radius of the circle is 6 units. The grid helps visualize and confirm this measurement, as the circle intersects the x-axis and y-axis at the points (6, 0), (-6, 0), (0, 6), and (0, -6).

### Solution:
The standard equation of a circle centered at the origin (h, k) with radius r is given by:

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

Since the center of the circle is at the origin (0, 0), h and k are both 0. The radius r is 6.

Substituting these values into the standard equation, we get:

\[
(x - 0)^2 + (y - 0)^2 = 6^2
\]

Simplifying, this becomes:

\[
x^2 + y^2 = 36
\]

**Therefore, the standard equation of the circle is:**

\[
x^2 + y^2 = 36
\]
Transcribed Image Text:**Problem 6) Write the standard equation of the circle.** **Explanation:** The diagram presents a circle centered at the origin (0, 0) on a coordinate plane. The radius of the circle is 6 units. The grid helps visualize and confirm this measurement, as the circle intersects the x-axis and y-axis at the points (6, 0), (-6, 0), (0, 6), and (0, -6). ### Solution: The standard equation of a circle centered at the origin (h, k) with radius r is given by: \[ (x - h)^2 + (y - k)^2 = r^2 \] Since the center of the circle is at the origin (0, 0), h and k are both 0. The radius r is 6. Substituting these values into the standard equation, we get: \[ (x - 0)^2 + (y - 0)^2 = 6^2 \] Simplifying, this becomes: \[ x^2 + y^2 = 36 \] **Therefore, the standard equation of the circle is:** \[ x^2 + y^2 = 36 \]
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