document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=35572>Question 35572</A>: <I><SPAN STYLE=\"word-break: break-all;\"> thanks for previous solution.  Can someone pls have a look at the following problems for me. 'Hope I'm in the right area. 'Am using a HP graphics calculator.\r<BR>\n" );
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document.write( "1. (1,-5) is one end of the diameter of a circle and the tangent through the point at the other end of the diameter has equation x+2y=11<BR>\n" );
document.write( "Find:<BR>\n" );
document.write( "a.  the equation of the diameter<BR>\n" );
document.write( "b.  the coordinates of the other end of the diameter<BR>\n" );
document.write( "c.  the coordinates of the centre of the circle\r<BR>\n" );
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document.write( "2. A base for jet fighters is located at F(4,1) The fighters have an operational range of 700kms.  An enemy bomber B at (0,-8) is heading for a target T at (16,0).  <BR>\n" );
document.write( "Given that 1 grid unit represents 100kms find:<BR>\n" );
document.write( "a.  the equation of the line representing the bomber's flight path from B to T<BR>\n" );
document.write( "b.  the equation of the boundary of the region which represents the range of the   fighters<BR>\n" );
document.write( "c.  the grid references of the points where the fighters can firstly & finally intercept the bomber\r<BR>\n" );
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document.write( "Thank you </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #21706 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=21706\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1. (1,-5)....POINT A SAY.... is one end of the diameter of a circle and the tangent through the point at the other end of the diameter...B SAY... has equation x+2y=11....OR....Y=(-1/2)*X+11/2...LINE L SAY<BR>\n" );
document.write( "Find:<BR>\n" );
document.write( "a. the equation of the diameter<BR>\n" );
document.write( "TANGENT IS PERPENDICULAR TO DIAMETER.HENCE AB IS PERPENDICLAR TO LINE L \r<BR>\n" );
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document.write( "SLOPE OF L IS -1/2 ...HENCE SLOPE OF AB THE PERPENDICULAR IS 2..HENCE <BR>\n" );
document.write( "HENCE EQN.OF AB THE DIAMETER IS <BR>\n" );
document.write( "Y+5=2(X-1)=2X-2<BR>\n" );
document.write( "Y=2X-7<BR>\n" );
document.write( "b. the coordinates of the other end of the diameter<BR>\n" );
document.write( "IF B IS (H,K)..THEN SLOPE OF AB =(K+5)/(H-1)<BR>\n" );
document.write( "(K+5)/(H-1)=2<BR>\n" );
document.write( "2H-2=K+5<BR>\n" );
document.write( "2H-K=7....................................I<BR>\n" );
document.write( "B IS ON TANGENT L...HENCE<BR>\n" );
document.write( "H+2K=11................................II<BR>\n" );
document.write( "EQN.II*2-EQN.I...GIVES<BR>\n" );
document.write( "2H+4K-2H+K=22-7=15<BR>\n" );
document.write( "5K=15<BR>\n" );
document.write( "K=3<BR>\n" );
document.write( "H=11-2*3=5<BR>\n" );
document.write( "SO B THE OTHER END OF DIAMETER IS (5,3)<BR>\n" );
document.write( "c. the coordinates of the centre of the circle<BR>\n" );
document.write( "IT IS MID POINT OF AB. HENCE CENTRE IS <BR>\n" );
document.write( "(1+5)/2 AND (-5+3)/2...OR....(3,-1)\r<BR>\n" );
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document.write( "2. A base for jet fighters is located at F(4,1) The fighters have an operational range of 700kms. An enemy bomber B at (0,-8) is heading for a target T at (16,0).<BR>\n" );
document.write( "Given that 1 grid unit represents 100kms find:<BR>\n" );
document.write( "a. the equation of the line representing the bomber's flight path from B to T<BR>\n" );
document.write( "EQN.OF BT ...TAKING SCALE OF 1 UNIT =100 KM...IS..USING (Y-Y1)/(Y2-Y1)=(X-X1)/X2-X1) FORMULA <BR>\n" );
document.write( "(Y+800)/(0+800)=(X)/(1600)<BR>\n" );
document.write( "Y+800=X/2.....OR X=2Y+1600................I<BR>\n" );
document.write( "b. the equation of the boundary of the region which represents the range of the fighters<BR>\n" );
document.write( "THE FIGHTER RANGE BEING 700 KM .FROM BASE AT (400,100) IS A CIRCLE WITH EQN.<BR>\n" );
document.write( "(X-H)^2+(Y-K)^2=R^2...SO...<BR>\n" );
document.write( "(X-400)^2+(Y-100)^2=700^2=490000...........................II\r<BR>\n" );
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document.write( "c. the grid references of the points where the fighters can firstly & finally intercept the bomber...THE INTERCEPTION TAKES PLACE AT A POINT WHERE LINE BT....EQN.I... CUTS THE CIRCLE EQN. II<BR>\n" );
document.write( "(2Y+1600-400)^2+(Y-100)^2=490000<BR>\n" );
document.write( "4Y^2+4800Y+1200^2+Y^2-200Y+100^2=490000<BR>\n" );
document.write( "5Y^2+4600Y+960000=0<BR>\n" );
document.write( "Y^2+920Y+192000=0<BR>\n" );
document.write( "Y^2+320Y+600Y+192000=0<BR>\n" );
document.write( "Y(Y+320)+600(Y+320)=0<BR>\n" );
document.write( "(Y+600)(Y+320)=0<BR>\n" );
document.write( "Y=-600....CORRESPONDINGLY X=2Y+1600=-2*600+1600=400...AND<BR>\n" );
document.write( "Y=-320....CORREPONDINGLY  X=-2*320+1600=960<BR>\n" );
document.write( "HENCE THE FIRST POINT OF INTERCEPTION COULD BE (400,-600)..OR GRID POINT (4,-6)<BR>\n" );
document.write( "AND THE FINAL POINT OF INTERCEPTION COULD BE (960,-320)..OR GRID POINT (9.6,-3.2)\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+graph%28+600%2C+600%2C+-500%2C+1500%2C+-1000%2C+1000%2C+%28x%2F2%29-800%2C100%2B%28490000-%28x-400%29%5E2%29%5E0.5%2C100-%28490000-%28x-400%29%5E2%29%5E0.5%29+ ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Tank you </SPAN>\n" );
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