SOLUTION: 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. 1. (1,-5) is one en

Question 35572: 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.
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
Find:
a. the equation of the diameter
b. the coordinates of the other end of the diameter
c. the coordinates of the centre of the circle

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).
Given that 1 grid unit represents 100kms find:
a. the equation of the line representing the bomber's flight path from B to T
b. the equation of the boundary of the region which represents the range of the fighters
c. the grid references of the points where the fighters can firstly & finally intercept the bomber

Thank you
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
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
Find:
a. the equation of the diameter
TANGENT IS PERPENDICULAR TO DIAMETER.HENCE AB IS PERPENDICLAR TO LINE L
SLOPE OF L IS -1/2 ...HENCE SLOPE OF AB THE PERPENDICULAR IS 2..HENCE
HENCE EQN.OF AB THE DIAMETER IS
Y+5=2(X-1)=2X-2
Y=2X-7
b. the coordinates of the other end of the diameter
IF B IS (H,K)..THEN SLOPE OF AB =(K+5)/(H-1)
(K+5)/(H-1)=2
2H-2=K+5
2H-K=7....................................I
B IS ON TANGENT L...HENCE
H+2K=11................................II
EQN.II*2-EQN.I...GIVES
2H+4K-2H+K=22-7=15
5K=15
K=3
H=11-2*3=5
SO B THE OTHER END OF DIAMETER IS (5,3)
c. the coordinates of the centre of the circle
IT IS MID POINT OF AB. HENCE CENTRE IS
(1+5)/2 AND (-5+3)/2...OR....(3,-1)
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).
Given that 1 grid unit represents 100kms find:
a. the equation of the line representing the bomber's flight path from B to T
EQN.OF BT ...TAKING SCALE OF 1 UNIT =100 KM...IS..USING (Y-Y1)/(Y2-Y1)=(X-X1)/X2-X1) FORMULA
(Y+800)/(0+800)=(X)/(1600)
Y+800=X/2.....OR X=2Y+1600................I
b. the equation of the boundary of the region which represents the range of the fighters
THE FIGHTER RANGE BEING 700 KM .FROM BASE AT (400,100) IS A CIRCLE WITH EQN.
(X-H)^2+(Y-K)^2=R^2...SO...
(X-400)^2+(Y-100)^2=700^2=490000...........................II
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
(2Y+1600-400)^2+(Y-100)^2=490000
4Y^2+4800Y+1200^2+Y^2-200Y+100^2=490000
5Y^2+4600Y+960000=0
Y^2+920Y+192000=0
Y^2+320Y+600Y+192000=0
Y(Y+320)+600(Y+320)=0
(Y+600)(Y+320)=0
Y=-600....CORRESPONDINGLY X=2Y+1600=-2*600+1600=400...AND
Y=-320....CORREPONDINGLY X=-2*320+1600=960
HENCE THE FIRST POINT OF INTERCEPTION COULD BE (400,-600)..OR GRID POINT (4,-6)
AND THE FINAL POINT OF INTERCEPTION COULD BE (960,-320)..OR GRID POINT (9.6,-3.2)

Tank you
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