document.write( "Question 1086213: What is the equation of the circle passing through (6,2) and tangent to the line x-4y-15=0 at (3,-3). \n" ); document.write( "

Algebra.Com's Answer #700374 by htmentor(1343) $\"\"$ $\"About$

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The center of the circle lies on the line perpendicular to the tangent line and passing through (3,-3).
\n" ); document.write( "Putting x-4y-15=0 in standard form we have y = x/4 - 15/4
\n" ); document.write( "Thus the equation of the line passing through the center is:
\n" ); document.write( "y + 3 = -4(x - 3) -> y = -4x + 9
\n" ); document.write( "The distance from the center of the circle to each of the two points is equal to the radius of the circle, by definition.
\n" ); document.write( "Let the center of the circle be the point (a,b)
\n" ); document.write( "Therefore sqrt((a-3)^2 + (b+3)^2) = sqrt((a-6)^2 + (b-2)^2)
\n" ); document.write( "And since the center lies on y = -4x + 9, we can substitute for b:
\n" ); document.write( "b = -4a + 9
\n" ); document.write( "Making the substitution, squaring both sides and collecting terms we are left with:
\n" ); document.write( "34a = 68 or a = 2
\n" ); document.write( "Therefore b = -4*2 + 9 = 1
\n" ); document.write( "So the center is (2,1) and the radius R = sqrt((2-3)^2+(1+3)^2) = sqrt(17)
\n" ); document.write( "Thus the equation of the circle is (x-2)^2 + (y-1)^2 = 17\r
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