document.write( "Question 331946: A circle is tangent to the y-axis and has a radius of 3 units.The center of the circle is in the third quadrant and lies on the graph of y - 2x = 0.What is the product of the coordinates (h, k) of the center of the circle? \n" ); document.write( "

Algebra.Com's Answer #237931 by galactus(183) $\"\"$ $\"About$

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Given the conditions, we know that h=-3. That is, since the circle is tangent to the y-axis, lies in the third quadrant, and the radius is 3, the x-coordinate of the center must be at x=-3.\r
\n" ); document.write( "\n" ); document.write( "To find the y-coordinate, k, of the center use y=2x.\r
\n" ); document.write( "\n" ); document.write( "y=2(-3)=-6\r
\n" ); document.write( "\n" ); document.write( "The center coordinates are (-3,-6). The product is obviously (-6)(-3)=18.\r
\n" ); document.write( "\n" ); document.write( "The equation of the circle is $\"%28x%2B3%29%5E2%2B%28y%2B6%29%5E2=9\"$ \r
\n" ); document.write( "\n" ); document.write( "To graph this on most calculators, we would have to solve this for y in terms of x.\r
\n" ); document.write( "\n" ); document.write( "If you have a graphing calculator, graph:\r
\n" ); document.write( "\n" ); document.write( " $\"-%28sqrt%28-x%5E2-6x%29%2B6%29\"$ for the lower half of the circle\r
\n" ); document.write( "\n" ); document.write( "and\r
\n" ); document.write( "\n" ); document.write( " $\"sqrt%28-x%5E2-6x%29-6\"$ for the uppr half.\r
\n" ); document.write( "\n" ); document.write( "Graph y=2x as well and you can see it pass through the center of the circle.\r
\n" ); document.write( "\n" ); document.write( "Casio calcualators have a nice Conics menu that graphs conics in terms of there equations without solving for y. It's a nice feature.\r
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