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Select the correct answer. What is the general form of the equation of a circle with its center at (-2, 1) and passing through (-4, 1)? A. x2 + y2 − 4x + 2y + 1 = 0 B. x2 + y2 + 4x − 2y + 1 = 0 C. x2 + y2 + 4x − 2y + 9 = 0 D. x2 − y2 + 2x + y + 1 = 0

User Matt Baer
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2 Answers

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Answer and Step-by-step explanation:

Answer:

x^2 +4x + y^2 -2y +1 =0

Explanation:

First we need to find the radius

Since the y coordinate is the same, the radius is the difference in the x coordinate -2 - (-4) = -2+4 = 2

A circle can be written in the form

(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius

(x--2)^2 + (y-1)^2 = 2^2

(x+2)^2 + (y-1)^2 = 4

FOIL ing

x^2 +4x+4 + y^2 -2y +1 = 4

Combining like terms

x^2 +4x + y^2 -2y +5 -4 =0

x^2 +4x + y^2 -2y +1 =0

User Dharmesh
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3.8k points
4 votes

Answer:

x^2 +4x + y^2 -2y +1 =0

Explanation:

First we need to find the radius

Since the y coordinate is the same, the radius is the difference in the x coordinate -2 - (-4) = -2+4 = 2

A circle can be written in the form

(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius

(x--2)^2 + (y-1)^2 = 2^2

(x+2)^2 + (y-1)^2 = 4

FOIL ing

x^2 +4x+4 + y^2 -2y +1 = 4

Combining like terms

x^2 +4x + y^2 -2y +5 -4 =0

x^2 +4x + y^2 -2y +1 =0

User Carl Whalley
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3.8k points