document.write( "Question 990644: Find the equation of the circle whose centre lies on the line x = 2y, passes throught the point (2,5), and for which the lenght of the tangent from the origin is sqrt{ 7)}}. Please recomend any coordinate geometory textbook for me \n" ); document.write( "
Algebra.Com's Answer #612443 by anand429(138)\"\" \"About 
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Since centre lies on x=2y,
\n" ); document.write( "Let the centre be (2a,a)
\n" ); document.write( "Eqn of circle will be
\n" ); document.write( "\"%28x-2a%29%5E2+%2B+%28y-a%29%5E2+=+r%5E2\"
\n" ); document.write( "Since it passes through (2,5), putting it we get,
\n" ); document.write( "\"%282-2a%29%5E2+%2B+%285-a%29%5E2+=+r%5E2\"
\n" ); document.write( "=> \"4a%5E2-8a%2B4+%2B+a%5E2-10a%2B25+=+r%5E2\"
\n" ); document.write( "=> \"5a%5E2-18a%2B29+=+r%5E2\"----------------------------------------(i)
\n" ); document.write( "Now length of tangent from origin is sqrt(7)
\n" ); document.write( "So, using pythagoras theorem,
\n" ); document.write( " (Distance of centre from origin)^2 = (length of tangent)^2 + (radius)^2
\n" ); document.write( "=> \"%282a-0%29%5E2%2B+%28a-0%29%5E2+=+7+%2B+r%5E2\"
\n" ); document.write( "=> \"5a%5E2-7+=+r%5E2\"-----------------------------(ii)
\n" ); document.write( "Subtracting (i) from (ii),
\n" ); document.write( "\"18a-36+=+0\"
\n" ); document.write( "=> \"a=2\"
\n" ); document.write( "Putting back in eqn (ii),
\n" ); document.write( "\"r%5E2=13\"
\n" ); document.write( "Putting both values in eqn of circle we get,
\n" ); document.write( "\"%28x-4%29%5E2+%2B+%28y-2%29%5E2+=+13\"
\n" ); document.write( "=> \"x%5E2-8x%2By%5E2-4y%2B7=0\"\r
\n" ); document.write( "\n" ); document.write( "P.S. - \"The Elements of Coordinate Geometry\" by S. L. Loney
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