document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=275511>Question 275511</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the equation of the line tangent to the circle at the indicated point.<BR>\n" );
document.write( "x^2 + y^2 = 25    <BR>\n" );
document.write( "(3,4)\r<BR>\n" );
document.write( "\n" );
document.write( "I believe the circle has a center of (0,0) and a radius of 5. I do know that the point (3,4) is perpendicular to the radius of the circle.\r<BR>\n" );
document.write( "\n" );
document.write( "I am not sure how to plug the line tanget points into this and find an equation for that tangent line though.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #200932 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=200932\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the line tangent to the circle would be perpendicular to the radius of the circle at that point.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the circle of x^2 + y^2 = 25 has a radius of 5 units and the center of the circle is at the point (0,0).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "to graph the circle you solve for y.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "equation would be y = +/- sqrt(25-x^2) and would look like this on the graph.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28600%2C600%2C-10%2C10%2C-10%2C10%2Csqrt%2825-x%5E2%29%2C-sqrt%2825-x%5E2%29%29 ALIGN=MIDDLE  ALT=\"graph%28600%2C600%2C-10%2C10%2C-10%2C10%2Csqrt%2825-x%5E2%29%2C-sqrt%2825-x%5E2%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The equation of the radius intersecting the circle at the point (3,4) would be found as follows:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "let x1,y1 = 0,0<BR>\n" );
document.write( "let x2,y2 = 3,4\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "slope = y2-y1 / x2-x1 = 4/3\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "y-intercept would be at point (0,0).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "equation would be y = mx + b where m is the slope and b is the y intersect.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "this makes the equation of the radius y = (4/3)*x\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "add that line to the graph and it looks like this.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28600%2C600%2C-10%2C10%2C-10%2C10%2Csqrt%2825-x%5E2%29%2C-sqrt%2825-x%5E2%29%2C%284%2F3%29%2Ax%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "line tangent to the graph at the point (3,4) will be perpendicular to this line.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "that means that the line will have a slope that is a negative reciprocal to the radius at that point.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "slope of the radius = (4/3).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "negative reciprocal of (4/3) is equal to -(3/4).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the line perpendicular to the radius goes through the point (3,4), so the y-intercept would be found by taking the equation of y = -(3/4)x + b and substituting 4 for y and 3 for x to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "4 = -(3/4)*3 + b\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "solve for b to get b = 25/4\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the equation of the line is y = -(3/4)x + (25/4)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "add that line to the graph and it looks like this:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28600%2C600%2C-10%2C10%2C-10%2C10%2Csqrt%2825-x%5E2%29%2C-sqrt%2825-x%5E2%29%2C%284%2F3%29%2Ax%2C-%283%2F4%29%2Ax+%2B+%2825%2F4%29%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you can see that the line goes through the point (3,4) and is perpendiular to the radius of the circle that also goes through the point (3,4).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the equation of the line tangent to the circle at the point (3,4) is:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "y = -(3/4)*x + 25/4<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );