SOLUTION: I need to find the general form of the equation of a circle. The problem in the book is "find center at the point (-3,1) and tangent to the y-axis. Im taking a correspondence cours

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Question 117909This question is from textbook Algebra and Trigonometry (Sullivan)
: I need to find the general form of the equation of a circle. The problem in the book is "find center at the point (-3,1) and tangent to the y-axis. Im taking a correspondence course and am having difficulty with the tangent part of the problem. Cheers! This question is from textbook Algebra and Trigonometry (Sullivan)

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Find the equation of a circle tangent to the y-axis and whose center is at (-3, 1).
The standard form for a circle of radius r and center at (h, k) is given by:
%28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2
You are given the center (h, k) as (-3, 1), so h = -3, and k = 1
Now for the radius, r:
Any line that is tangent to (touching) a circle is perpendicular to the radius of the circle.
Since the given circle is tangent to (touching) the y-axis, you know that the radius of the circle is perpendicular to the y-axis and you also know that center is at (-3, 1), so the radius is 3.
Now you can write the equation:
%28x-%28-3%29%29%5E2+%2B+%28y-1%29%5E2+=+3%5E2 Simplify this to:
%28x%2B3%29%5E2+%2B+%28y-1%29%5E2+=+9