document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1207940>Question 1207940</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The tangent line to a circle may be defined as the line that intersects the circle in a single point, called the point of tangency. If the equation of the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b, show that:\r<BR>\n" );
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document.write( "A. r^2(1 + m^2) = b^2\r<BR>\n" );
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document.write( "B. The point of tangency is [(-r^2 m)/b, (r^2/b)]\r<BR>\n" );
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document.write( "C. The tangent line is perpendicular to the line containing the center of the circle and point of tangency. \r<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 #846104 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=846104\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <BR>\n" );
document.write( "Since tangent and circle meet at only one point there is only one solution . Let the point be (x,y)<BR>\n" );
document.write( "The equation of the circle is given<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2%2By%5E2=r%5E2 ALIGN=MIDDLE  ALT=\"x%5E2%2By%5E2=r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "The equation of the tangent line is given<BR>\n" );
document.write( "y=mx+b<BR>\n" );
document.write( "substitute y= mx+b in the equation of circle<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%5E2+%2B+%28mx%2Bb%29%5E2+=+r%5E2 ALIGN=MIDDLE  ALT=\"+x%5E2+%2B+%28mx%2Bb%29%5E2+=+r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%5E2+%2B+m%5E2x%5E2%2B2mbx+%2Bb%5E2+=+r%5E2 ALIGN=MIDDLE  ALT=\"+x%5E2+%2B+m%5E2x%5E2%2B2mbx+%2Bb%5E2+=+r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28m%5E2%2B1%29x%5E2%2B+2mbx+%2B%28b%5E2-r%5E2%29=0 ALIGN=MIDDLE  ALT=\"%28m%5E2%2B1%29x%5E2%2B+2mbx+%2B%28b%5E2-r%5E2%29=0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "This is a quadratic equation and since tangent and circle meet at only one point there is only one solution<BR>\n" );
document.write( "Discriminant <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2-4ac+=0 ALIGN=MIDDLE  ALT=\"b%5E2-4ac+=0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Substitute b ,a and c  from the above equation .<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%282mb%29%5E2++-++4%28%28m%5E2%2B1%29%28b%5E2-r%5E2%29%29=0 ALIGN=MIDDLE  ALT=\"%282mb%29%5E2++-++4%28%28m%5E2%2B1%29%28b%5E2-r%5E2%29%29=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+4m%5E2b%5E2-4%28m%5E2b%5E2-m%5E2r%5E2%2Bb%5E2-r%5E2%29=0 ALIGN=MIDDLE  ALT=\"+4m%5E2b%5E2-4%28m%5E2b%5E2-m%5E2r%5E2%2Bb%5E2-r%5E2%29=0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+4m%5E2b%5E2-4m%5E2b%5E2%2B4m%5E2r%5E2-4b%5E2%2B4r%5E2%29=0 ALIGN=MIDDLE  ALT=\"+4m%5E2b%5E2-4m%5E2b%5E2%2B4m%5E2r%5E2-4b%5E2%2B4r%5E2%29=0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+4m%5E2r%5E2-4b%5E2%2B4r%5E2=0 ALIGN=MIDDLE  ALT=\"+4m%5E2r%5E2-4b%5E2%2B4r%5E2=0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+4m%5E2r%5E2%2B4r%5E2=4b%5E2 ALIGN=MIDDLE  ALT=\"+4m%5E2r%5E2%2B4r%5E2=4b%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r%5E2%28m%5E2%2B1%29+=+b%5E2+ ALIGN=MIDDLE  ALT=\"r%5E2%28m%5E2%2B1%29+=+b%5E2+\"   DISABLED_style= \"max-width:90%; height:auto;\" >-----------------------------------------A<BR>\n" );
document.write( "B<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28m%5E2%2B1%29x%5E2%2B+2mbx+%2B%28b%5E2-r%5E2%29=0 ALIGN=MIDDLE  ALT=\"%28m%5E2%2B1%29x%5E2%2B+2mbx+%2B%28b%5E2-r%5E2%29=0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Since the discriminant is zero, the equation has exactly one solution for x. The solution for x in a quadratic equation is given by <BR>\n" );
document.write( "x = -b/2a<BR>\n" );
document.write( "Here it is -2mb/2(m^2+1) = -mb/(m^2+1)<BR>\n" );
document.write( "substitute this value of x into the  equation y=mx+b <BR>\n" );
document.write( "y = m(-mb/(m^2+1)) +b )<BR>\n" );
document.write( "y = (-m^2b/(m^2+1)) +b )<BR>\n" );
document.write( "y= (-m^2b +m^2b+b)/(m^2+1)<BR>\n" );
document.write( "y = b/(m^2+1)-----------------------------------------------1\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r%5E2%28m%5E2%2B1%29+=+b%5E2+%7D%7D++++from+A%0D%0A%7B%7B%7B%28m%5E2%2B1%29=+b%5E2%2Fr%5E2 ALIGN=MIDDLE  ALT=\"r%5E2%28m%5E2%2B1%29+=+b%5E2+%7D%7D++++from+A%0D%0A%7B%7B%7B%28m%5E2%2B1%29=+b%5E2%2Fr%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Substitute(m^2+1)  in 1<BR>\n" );
document.write( "y = b/(b^2/r^2)<BR>\n" );
document.write( "y=r^2/b<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+-mb%2F%28m%5E2%2B1%29+=+-mb%2F%28b%5E2%2Fr%5E2%29+=+-%28mr%5E2%2Fb%29 ALIGN=MIDDLE  ALT=\"x+=+-mb%2F%28m%5E2%2B1%29+=+-mb%2F%28b%5E2%2Fr%5E2%29+=+-%28mr%5E2%2Fb%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "The point of tangency is (-r^2 m)/b, (r^2/b) ……………………….B<BR>\n" );
document.write( "C<BR>\n" );
document.write( "We have to prove tangent and radius are perpendicular.<BR>\n" );
document.write( "From circle equation we know co ordinates of centre are (0,0)<BR>\n" );
document.write( "The point of contact of radius and tangent are  (-r^2 m)/b, (r^2/b)<BR>\n" );
document.write( "Find slope using two point formula <BR>\n" );
document.write( "Slope = ((r^2/b))/((-r^2m/b))<BR>\n" );
document.write( "Slope =(- 1/m)<BR>\n" );
document.write( "Slope of tangent line = m<BR>\n" );
document.write( "M*(-1/m)= -1<BR>\n" );
document.write( "Hence The tangent line is perpendicular to the line containing the center of the circle and point of tangency………….C\r<BR>\n" );
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