document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=64612>Question 64612</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the center, radius, and x- and y-intercepts of the circle given by the equation 3x2 + 3y2 - 6x + 5y = 0.   </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #46480 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=46480\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Find the center, radius, and x- and y-intercepts of the circle given by the equation 3x2 + 3y2 - 6x + 5y = 0.<BR>\n" );
document.write( "DIVIDING BY 3<BR>\n" );
document.write( "X^2-2X+Y^2+5Y/3=0<BR>\n" );
document.write( "(X-1)^2+[Y+(5/6)]^2=1+(5/6)^2=61/36<BR>\n" );
document.write( "HENCE CENTER = (1,-5/6)......RADIUS = (1/6)*SQRT(61)<BR>\n" );
document.write( "AT Y=0,WE GET<BR>\n" );
document.write( " X^2-2X=0=X(X-2)......HENCE X INTERCEPTS ARE 0 AND 2.<BR>\n" );
document.write( "AT X=0,WE GET<BR>\n" );
document.write( "Y^2+5Y/3=0=Y(Y+5/3)....HENCE Y INTERCEPTS ARE 0 AND -5/3 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );