document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1039212>Question 1039212</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I have another questuon pls.<BR>\n" );
document.write( "The diameter of a cirlce whose endpoints are (-3,-3) and (3,3)<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #653966 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=653966\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> if the end points of the diameter are (-3,-3) and (3,3), then the length of the diameter would be the length of the line segment between those 2 points.\r<BR>\n" );
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document.write( "you have:\r<BR>\n" );
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document.write( "(x1,y1) = (-3,-3)<BR>\n" );
document.write( "(x2,y2) = (3,3)\r<BR>\n" );
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document.write( "let d represent length of the diameter.\r<BR>\n" );
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document.write( "d = sqrt((x2-x1)^2 + (y2-y1)^2).\r<BR>\n" );
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document.write( "formula becomes d = sqrt((3 - (-3))^2 + (3 - (-3))^2).<BR>\n" );
document.write( "simplify this to get d = sqrt(3+3)^2 + (3+3)^2).<BR>\n" );
document.write( "simplify further to get d = sqrt(6^2 + 6^2).<BR>\n" );
document.write( "simplify further to get d = sqrt(36 + 36).<BR>\n" );
document.write( "simplify further to get d = sqrt(72).<BR>\n" );
document.write( "simplify further to get d = 6*sqrt(2).\r<BR>\n" );
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document.write( "the solution to your problem is that that the length of the diameter of the circle is 6 * sqrt(2).\r<BR>\n" );
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document.write( "if that's all you wanted, you can stop here.<BR>\n" );
document.write( "the rest is just more stuff if you want to learn more about this problem.\r<BR>\n" );
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document.write( "the general formula of a circle is (x-h)^2 + (y-k)^2 = r^2<BR>\n" );
document.write( "(h,k) is the center of the circle.<BR>\n" );
document.write( "r is the radius.\r<BR>\n" );
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document.write( "we got d = 6*sqrt(2).<BR>\n" );
document.write( "that's the length of the diameter.<BR>\n" );
document.write( "the length of the radius is half that, so the length of the radius would be 3*sqrt(2).<BR>\n" );
document.write( "that would be r.<BR>\n" );
document.write( "r^2 would be (3*sqrt(2))^2 = 9*2 = 18.\r<BR>\n" );
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document.write( "the center of your circle is at the midpoint of the line segment between (-3,-3) and (3,3).\r<BR>\n" );
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document.write( "the formula for midpoint is ((x1+x2)/2,(y1+y2)/2).<BR>\n" );
document.write( "that puts your midpoint at (0,0) which is the center of the circle.\r<BR>\n" );
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document.write( "the formula for your circle becomes x^2 + y^2 = 18.\r<BR>\n" );
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document.write( "that is shown in the following graph with the points of (-3,-3) and (3,3) displayed because they lie on the line of the equation of y = x.\r<BR>\n" );
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document.write( "how i derive that is based on the slope intercept form of the equation for a straight line that is y = mx + b, where m is the slope and b is the y-intercept.\r<BR>\n" );
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document.write( "the slope is (y2-y1)/(x2-x1) which becomes 6/6 which becomes 1.\r<BR>\n" );
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document.write( "the y-intercerpt is found by taking any one of the points and replacing x and y with them in the general equation and then solving for b.\r<BR>\n" );
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document.write( "y = mx + b becomes y = x + b which becomes 3 = 3 + b if we use the point (3,3).\r<BR>\n" );
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document.write( "solve for b to get b = 0.\r<BR>\n" );
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document.write( "the slope intercept form of the equation through the points (-3,-3) and (3,3) is y = x.\r<BR>\n" );
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document.write( "here's the graph.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2016/062502.jpg\" alt=\"$$$\" </>\r<BR>\n" );
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