document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1088896>Question 1088896</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Given circle x^2-2x+y^2-16y+39=0 with center A and y-intercepts B(0;p) and C(0;q) where p <<BR>\n" );
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document.write( "1 determine the values of p and q<BR>\n" );
document.write( "2 show that point D(2;13) lies on circumference of the circle<BR>\n" );
document.write( "3 prove that point B,A and D are collinear </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #703210 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=703210\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x^2-2x+y^2-16y+39=0 <BR>\n" );
document.write( "To get the equation of a circle, one needs to write this as (x-a)^2+(y-b)^2=r^2<BR>\n" );
document.write( "The first thing is to move the constant to the right side.<BR>\n" );
document.write( "Then rewrite as x^2-2x+y^2-16y=-39<BR>\n" );
document.write( "One needs to complete the square of x^2=2x.  That is half the -2, squared, which is +1; x^2-2x+1.  One has to add 1 to the other side to balance the equation.<BR>\n" );
document.write( "Do the same with the y, by taking half of -16, or -8, and squaring it to get 64.<BR>\n" );
document.write( "complete the square of x and y to get x^2-2x+1+y^2-16x+64=-39+1+64<BR>\n" );
document.write( "This can be written as the sum of two squares equalling a constant, which is the radius squared.<BR>\n" );
document.write( "(x-1)^2+(y-8)^2=26<BR>\n" );
document.write( "center is at (1, 8) and radius is sqrt(26). One takes the opposite sign of the x-1, and the center has x coordinate 1.  The y-coordinate of the center is 8<BR>\n" );
document.write( "y-intercept is where x=0, so one puts a 0 into any place there is an x.  One could do it at the beginning as well and get y^2-16y+39=0; that is y^2-16y=-39, and completing the square y^2-16y+64=25. That is (y-8)^2=5^2<BR>\n" );
document.write( "1+(y-8)^2=26<BR>\n" );
document.write( "(y-8)^2=25<BR>\n" );
document.write( "y=8+/- 5 or 3 and 13. p is 3 and q is 13.<BR>\n" );
document.write( "center is (1, 8) and B is (0, 3) and D is (2, 13).  If these points are collinear, the slope between any two of them must be the same. One does the slope formula for them.<BR>\n" );
document.write( "slope between center and B is -5/-1 or 5<BR>\n" );
document.write( "slope between B and D is 10/2=5<BR>\n" );
document.write( "check to see if (2, 13) is a solution of the circle equation by substituting (2, 13) into the equation for x and y respectively: 1^2+5^2=26, and that is true.<BR>\n" );
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