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Question 31 of 40 What is the equation of a circle with center (-2, 3) and radius 4? A. (x+2)? +(y - 3)? = 16 B. (x+2)? + (y - 3)? - 4 C. (x-2)? + (y +3)? = 16 D. (x+2)? - (y - 3)? = 16 %3D

Question 31 of 40 What is the equation of a circle with center (-2, 3) and radius 4? A. (x+2)? +(y - 3)? = 16 B. (x+2)? + (y - 3)? - 4 C. (x-2)? + (y +3)? = 16 D. (x+2)? - (y - 3)? = 16 %3D

Elementary Geometry For College Students, 7e
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
Problem 1CT
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**Question 31 of 40** **What is the equation of a circle with center (-2, 3) and radius 4?** - **A.** \((x + 2)^2 + (y - 3)^2 = 16\) - **B.** \((x + 2)^2 + (y - 3)^2 = 4\) - **C.** \((x - 2)^2 + (y + 3)^2 = 16\) - **D.** \((x + 2)^2 - (y - 3)^2 = 16\) The correct equation to use for a circle with a center at \((h, k)\) and radius \(r\) is \((x - h)^2 + (y - k)^2 = r^2\). Given the center \((-2, 3)\) and radius \(4\), substitute \(h = -2\), \(k = 3\), and \(r = 4\) into the equation: \((x + 2)^2 + (y - 3)^2 = 16\). Thus, the correct option is **A**.
Transcribed Image Text:**Question 31 of 40** **What is the equation of a circle with center (-2, 3) and radius 4?** - **A.** \((x + 2)^2 + (y - 3)^2 = 16\) - **B.** \((x + 2)^2 + (y - 3)^2 = 4\) - **C.** \((x - 2)^2 + (y + 3)^2 = 16\) - **D.** \((x + 2)^2 - (y - 3)^2 = 16\) The correct equation to use for a circle with a center at \((h, k)\) and radius \(r\) is \((x - h)^2 + (y - k)^2 = r^2\). Given the center \((-2, 3)\) and radius \(4\), substitute \(h = -2\), \(k = 3\), and \(r = 4\) into the equation: \((x + 2)^2 + (y - 3)^2 = 16\). Thus, the correct option is **A**.
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