document.write( "Question 117909This question is from textbook Algebra and Trigonometry (Sullivan)
\n" ); document.write( ": 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! \n" ); document.write( "
Algebra.Com's Answer #85893 by Earlsdon(6294)\"\" \"About 
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Find the equation of a circle tangent to the y-axis and whose center is at (-3, 1).\r
\n" ); document.write( "\n" ); document.write( "The standard form for a circle of radius r and center at (h, k) is given by:
\n" ); document.write( "\"%28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2\"
\n" ); document.write( "You are given the center (h, k) as (-3, 1), so h = -3, and k = 1\r
\n" ); document.write( "\n" ); document.write( "Now for the radius, r:
\n" ); document.write( "Any line that is tangent to (touching) a circle is perpendicular to the radius of the circle.
\n" ); document.write( "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.
\n" ); document.write( "Now you can write the equation:
\n" ); document.write( "\"%28x-%28-3%29%29%5E2+%2B+%28y-1%29%5E2+=+3%5E2\" Simplify this to:
\n" ); document.write( "\"%28x%2B3%29%5E2+%2B+%28y-1%29%5E2+=+9\"
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