document.write( "Question 1209260: Find the equation of the tangents drawn from the point (4, 7) to the circle:
\n" ); document.write( " (x - 2)² + y² + 4y = 0.\r
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Algebra.Com's Answer #847929 by mccravyedwin(405) $\"\"$ $\"About$

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document.write( "Complete the square on the y-terms by adding 4 to both sides.\r\n" );
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document.write( "Factor the trinomial:\r\n" );
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document.write( "Compare to the standard form of a circle:\r\n" );
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document.write( "And we see that the center of the circle is (h,k) = (2,-2)\r\n" );
document.write( "and the radius is r=2\r\n" );
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document.write( "So we sketch the figure:\r\n" );
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document.write( "Let X and Q be the points of tangency, we draw radii OX and OQ,\r\n" );
document.write( "then we draw OP and draw XY parallel to the x-axis.\r\n" );
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document.write( "We want the equations of PX and PQ.\r\n" );
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document.write( "We now see that the equation of the tangent PQ \r\n" );
document.write( "is the vertical line x = 4. \r\n" );
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document.write( "We also see that PQ is 9 units in length, 7 units above the x-axis\r\n" );
document.write( "and 2 units below the x-axis.\r\n" );
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document.write( "From right triangle OPQ, we see that\r\n" );
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document.write( "We use the identity \r\n" );
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document.write( "Since , by right triangle PXY,\r\n" );
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document.write( " and  are complementary and\r\n" );
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document.write( "Finally we use the point-slope formula to find the equation of tangent PX\r\n" );
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document.write( "That's the equation of the tangent PX\r\n" );
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document.write( "and the equation of the tangent PQ is the vertical line x = 4.\r\n" );
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document.write( "Edwin

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