document.write( "Question 192952: If (4,5) and (6,1) are the endpoints of the diameter, write the equation of this circle. i dont know what formulas to use or how to actually solve this problem \n" ); document.write( "
Algebra.Com's Answer #144805 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Step 1: Find the center of the circle\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "To find the center of the circle, you need to find the midpoint of the diameter with (4,5) and (6,1) as the endpoints. So simply average the corresponding coordinates to get:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"x%5Bmid%5D=%284%2B6%29%2F2=10%2F2=5\". So the x-coordinate of the center is \"x=5\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"y%5Bmid%5D=%281%2B5%29%2F2=6%2F2=3\". So the y-coordinate of the center is \"y=3\"\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So the center is (5,3). Since the center of \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2\" \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "is (h,k), this means that \"h=5\" and \"k=3\"\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "-----------------------------------------------------------------\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Step 2: Finding the radius\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "To find the radius, we need to find length of the diameter. To find the diameter's length, we need to find the distance from (4,5) to (6,1). To do that, we can use the distance formula:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"d=sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2%2B%28y%5B1%5D-y%5B2%5D%29%5E2%29\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Note: The point is (4,5) which means that \"x%5B1%5D=4\" and \"y%5B1%5D=5\". Similarly, the point is (6,1) which means that \"x%5B2%5D=6\" and \"y%5B2%5D=1\".\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"d=sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2%2B%28y%5B1%5D-y%5B2%5D%29%5E2%29\" Start with the distance formula.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"d=sqrt%28%284-6%29%5E2%2B%285-1%29%5E2%29\" Plug in \"x%5B1%5D=4\", \"x%5B2%5D=6\", \"y%5B1%5D=5\", and \"y%5B2%5D=1\".\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"d=sqrt%28%28-2%29%5E2%2B%285-1%29%5E2%29\" Subtract \"6\" from \"4\" to get \"-2\".\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"d=sqrt%28%28-2%29%5E2%2B%284%29%5E2%29\" Subtract \"1\" from \"5\" to get \"4\".\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"d=sqrt%284%2B%284%29%5E2%29\" Square \"-2\" to get \"4\".\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"d=sqrt%284%2B16%29\" Square \"4\" to get \"16\".\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"d=sqrt%2820%29\" Add \"4\" to \"16\" to get \"20\".\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"d=sqrt%284%2A5%29\" Factor \"20\" to get \"4%2A5\".\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"d=sqrt%284%29%2Asqrt%285%29\" Break up the square root.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"d=2%2Asqrt%285%29\" Take the square root of \"4\" to get \"2\".\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So the diameter is \"2%2Asqrt%285%29\" units long. Take half of this length to get the radius:\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"r=%282%2Asqrt%285%29%29%2F2=%28cross%282%29%2Asqrt%285%29%29%2Fcross%282%29=sqrt%285%29\"\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So \"r=sqrt%285%29\" which means that the radius is \"sqrt%285%29\" units long\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Since \"r=sqrt%285%29\", this means that \"r%5E2=%28sqrt%285%29%29%5E2=5\" or just \"r%5E2=5\"\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2\" Start with the general formula for a circle\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"%28x-5%29%5E2%2B%28y-3%29%5E2=5\" Plug in \"h=5\" and \"k=3\" and \"r%5E2=5\"\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So the equation of the circle with the points (4,5) and (6,1) as the endpoints of the diameter is \"%28x-5%29%5E2%2B%28y-3%29%5E2=5\"
\n" ); document.write( "
\n" ); document.write( "
\n" );