document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=990689>Question 990689</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please my good genius Algebra.Com help me with this !Find the equation of the tangent to the circle x^2 + y^2 + ax + 2ay = 3 at the point (0, b) in terms of a and b. If this tangent has a gradient of 1/4,find the relation between a and b. Hence, find the equations of the two possible circles. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #612432 by </B><A HREF=/tutors/aboutme.mpl?userid=anand429><B>anand429(138)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=anand429><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=anand429><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=612432\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Equation of circle is<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%5E2+%2B+y%5E2+%2B+ax+%2B+2ay+=+3 ALIGN=MIDDLE  ALT=\"+x%5E2+%2B+y%5E2+%2B+ax+%2B+2ay+=+3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "=> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%2Ba%2F2%29%5E2+-a%5E2%2F4+%2B+%28y%2Ba%29%5E2+-a%5E2+=3 ALIGN=MIDDLE  ALT=\"%28x%2Ba%2F2%29%5E2+-a%5E2%2F4+%2B+%28y%2Ba%29%5E2+-a%5E2+=3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "=> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%2Ba%2F2%29%5E2+%2B+%28y%2Ba%29%5E2+=+3%2B5a%5E2%2F4 ALIGN=MIDDLE  ALT=\"%28x%2Ba%2F2%29%5E2+%2B+%28y%2Ba%29%5E2+=+3%2B5a%5E2%2F4\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "So the center of circle is<BR>\n" );
document.write( "(-a/2, -a) and radius is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%283%2B5a%5E2%2F4%29 ALIGN=MIDDLE  ALT=\"sqrt%283%2B5a%5E2%2F4%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "\n" );
document.write( "Slope of radius at (0,b) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-a-b%29%2F%28-a%2F2+-0%29 ALIGN=MIDDLE  ALT=\"%28-a-b%29%2F%28-a%2F2+-0%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "=2(a+b)/a<BR>\n" );
document.write( " SO slope of tangent at (0,b) = (-a/(2(a+b))) (Since radius and tangent are perpendicular and so product of their slopes will be -1)\r<BR>\n" );
document.write( "\n" );
document.write( "So let equation of tangent be <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+%28-a%2F%282%28a%2Bb%29%29%29x+%2B+c ALIGN=MIDDLE  ALT=\"+y+=+%28-a%2F%282%28a%2Bb%29%29%29x+%2B+c\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Since it passes through (0,b)<BR>\n" );
document.write( "So,<BR>\n" );
document.write( "b=(-a/(2(a+b)))*0 + c<BR>\n" );
document.write( "=> b=c<BR>\n" );
document.write( " So the equation of tangent becomes,<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+%28-a%2F%282%28a%2Bb%29%29%29x+%2B+b ALIGN=MIDDLE  ALT=\"+y+=+%28-a%2F%282%28a%2Bb%29%29%29x+%2B+b\"   DISABLED_style= \"max-width:90%; height:auto;\" > --------------Required equation of tangent<BR>\n" );
document.write( " If this tangent has a gradient of 1/4<BR>\n" );
document.write( "then, <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-a%2F%282%28a%2Bb%29%29%29+=+1%2F4 ALIGN=MIDDLE  ALT=\"%28-a%2F%282%28a%2Bb%29%29%29+=+1%2F4\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "=> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-2a=a%2Bb ALIGN=MIDDLE  ALT=\"-2a=a%2Bb\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "=> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3a%2Bb=0 ALIGN=MIDDLE  ALT=\"3a%2Bb=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >  ----(i) ----------------Required relation between a and b\r<BR>\n" );
document.write( "\n" );
document.write( "Also, since (0,b) lies on circle,So<BR>\n" );
document.write( "0^2 + b^2 + a*0 + 2ab =3<BR>\n" );
document.write( "=> b^2 + 2ab - 3 = 0<BR>\n" );
document.write( "Putting value of b from eqn. (i) above,<BR>\n" );
document.write( "9a^2 -6a^2 - 3 =0<BR>\n" );
document.write( "=> a^2 =1<BR>\n" );
document.write( "=> a=1 or a = -1<BR>\n" );
document.write( "Putting these values of a in original equations of circle, we get<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2+%2B+y%5E2+%2B+x+%2B+2y+=+3 ALIGN=MIDDLE  ALT=\"x%5E2+%2B+y%5E2+%2B+x+%2B+2y+=+3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "and<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2+%2B+y%5E2+-+x+-+2y+=+3 ALIGN=MIDDLE  ALT=\"x%5E2+%2B+y%5E2+-+x+-+2y+=+3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( " as two equations of possible circles. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );