# What is the general equation of a circle centered at the point (h, k)

Anonymous
timer Asked: Dec 24th, 2017
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Question description

There are free response questions that must show work on the free response....but for the multiple choice the correct answer is suffucient

Aljon2017
School: Purdue University

Hi, please see attached file for the solution.Please check if I got everything covered.

QUESTION 1

What is the general equation of a circle centered at the point (h, k)?
The general equation of a circle with center at (h, k) and radius r can be computed by first determining its
center-radius form which is:
(x - h)2 + (y - k)2 = r2
x2 – 2hx +h2 +y2 – 2ky + k2 = r2
x2 + y2 – 2hx – 2ky +(h2 + k2 - r2) = 0
3 points

QUESTION 2
What is the equation of the circle with center at (2, 4) and radius of 6?
h=2
k=4
r=6
(x – 2)2 + (y – 4)2 = 62
x2 – 4x + 4 + y2 – 8y + 16 = 36
x2 – 4x + y2 – 8y –16 = 0
3 points

QUESTION 3

What is the general equation of a circle centered at the origin?
h=0
k=0
r=r
(x - 0)2 + (y – 0)2 = r2
x2 + y2 – r2 = 0

3 points

QUESTION 4

Find the equation of the circle with center at (3, 2) and through the point (5, 4).
The radius can be computed using the distance formula:
r = √(𝑦2 − 𝑦1 )2 + (𝑥2 − 𝑥1 )2 = √(4 − 2)2 + (5 − 3)2 = 2√2
h=3
k=2
x2 + y2 – 2hx – 2ky +(h2 + k2 - r2) = 0
x2 + y2 – 2(3) x – 2(2)y +(32 + 22 – (2√2)2) = 0
x2 + y2 – 6x – 4y + 21 = 0

3 points

QUESTION 5

If a circle has a center at the point (3, 4) and goes through the point (6, 8), what is the radius?
The radius can be computed using the distance formula:
r = √(𝑦2 − 𝑦1 )2 + (𝑥2 − 𝑥1 )2 = √(8 − 4)2 + (6 − 3)2 = 5

3 points

QUESTION 6

Find the equation of the circle with center at (-3, 1) and through the point (2, 13).
The radius can be computed using the distance formula:
r = √(𝑦2 − 𝑦1 )2 + (𝑥2 − 𝑥1 )2 = √(13 − 1)2 + (2 − (−3))2 = 13
h=-3
k=1
x...

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Anonymous
Excellent job

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