document.write( "Question 1156566: A curve is traced by a point (x,y) which moves such that its distance from the point A(-1,1) is three times its distance from the point B(2,-1). Determine the equation of the curve. \n" ); document.write( "
Algebra.Com's Answer #779295 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "The distance from (x,y) to A(-1,1) is \"sqrt%28%28x%2B1%29%5E2%2B%28y-1%29%5E2%29\".

\n" ); document.write( "The distance from (x,y) to B(2,-1) is \"sqrt%28%28x-2%29%5E2%2B%28y%2B1%29%5E2%29\".

\n" ); document.write( "The distance from A is 3 times the distance from B:

\n" ); document.write( "\"sqrt%28%28x%2B1%29%5E2%2B%28y-1%29%5E2%29+=+3%28sqrt%28%28x-2%29%5E2%2B%28y%2B1%29%5E2%29%29\"

\n" ); document.write( "The square of the distance from A is 9 times the square of the distance from B:

\n" ); document.write( "\"%28x%2B1%29%5E2%2B%28y-1%29%5E2+=+9%28%28x-2%29%5E2%2B%28y%2B1%29%5E2%29\"

\n" ); document.write( "I started down that path, and the numbers got ugly; so I decided to try another path to the answer.

\n" ); document.write( "Without doing any detailed calculations with the above equation, we can see that the equation is going to be the equation of a circle (there are going to be x^2 and y^2 terms with the same coefficient).

\n" ); document.write( "There are two points on the line containing A and B that satisfy the condition that the distance from A is 3 times the distance from B. Those two points will be the endpoints of a diameter of the circle.

\n" ); document.write( "The point between A and B is 3/4 of the way from A to B; that is (5/4,-1/2).

\n" ); document.write( "The point to the right of B is such that its distance from B is twice the distance from A to B; that point is (7/2,-2).

\n" ); document.write( "The center of the circle is the midpoint of that diameter: (19/8,-5/4).

\n" ); document.write( "The equation is of the form

\n" ); document.write( "\"%28x-19%2F8%29%5E2%2B%28y%2B5%2F4%29%5E2+=+r%5E2\"

\n" ); document.write( "The square of the radius of the circle is the square of the distance from the center to either endpoint of the diameter: \"%289%2F8%29%5E2%2B%283%2F4%29%5E2+=+117%2F64\"

\n" ); document.write( "The equation is

\n" ); document.write( "\"%28x-19%2F8%29%5E2%2B%28y%2B5%2F4%29%5E2+=+117%2F64\"

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