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Lecture 7 [16 March, 2020] MATH 127 (Prof Farid)

  

 MATH -127 (Coordinate Geometry)

Chapter -3
The Circle



    • The pole is outside the circle.
    • The pole is inside the circle 
    • The pole is on the polar 

♦ Polar and Pole: If through a point P, we draw a straight line to meet a circle or conic at two points Q and R, the locus of the points of intersection of the tangents drawn at Q, R is a line known as the polar and the point P is the pole of the polar.
locus is the line that is made of a 'set of points'.

    Question 1: Find the pole of the line  2xy=6   with respect to the circle  5x2+5y2=9  
Solution: given that, 
  2xy=6  
  5x2+5y2=9  
    The equation of the polar with respect to the circle,  5x2+5y2=9 of P(x1,y1) is  5xx1+5yy1=9  
Since equation  (1.3) is identical with the given line, we have,
  5x12=5y12=96x1=35,     y1=310  
The required pole is  (35,310)  

    Question 2: Show that the equation to the pair of tangents drawn from the origin to the circle  x2+y2+2gx+2fy+c=0 is  c(x2+y2)=(gx+fy)2  
Solution The equation to the pair of tangents is,
  SS1=T2(x2+y2+2gx+2fy+c)c=[x×0+y×0+g(x+0)+f(y+0)+c]2c(x2+y2)+c(2gx+2fy+c)=(gx+fy+c)2+(gx+fy)2+2(gx+fy)c+c2c(c2+y2)=(gx+fy)2  


Question 3: Prove that if the polar of a point P with respect to the circle  x2+y2=37 touches the circle  (x3)2+(y+2)2=25 then the locus of p must be a cone.
Solution: Given circles are  x2+y2=37 (x3)2+(y+2)2=25  
Let, P(x1,y1) be the pole.
Then the polar of P with respect to the circle  x2+y2=37 is  xx1+yy1=37  
Since (3.1) touches (3.2) we have, 
  3x12y137x12+y12  
Since length of the perpendicular from (3,2) to (3.1) is equal to the radius of the circle. so that,
  (3x12y137)2=25(x12+y12)16x12+21y12+12xy+222x1148y11369=0  
the required locus is,    16x12+21y12+12xy+222x1148y11369=0  


    Question 4: Find the pole of the straight line  9x+y=28 with respect to the circle  2x2+2y23x+5y7=0  
Solution: The given equations are 
  9x+y=28 2x2+2y23x+5y7=0  
If (x1,y1) be the required point the line (4.1) must coincide with the polar of (x1,y1) whose equation is,  2xx1+2yy132(x1+x)+52(y1+y)7=0x(4x13)+y(4y1+5)3x1+4y114=0  
Since (4.1) and (4.3) are the same, we have 
  4x139=4y1+51=3x1+5y11428  
Hence,  x1=9y1+143x1117y1=126  
solving equation (4.4) and (4.5) we get, x1=3,  y1=1
The required point is (3,1)


    Question 5: Find the polar of the point (4,1) with respect to the circle  2x2+2y2=11  
Solution: The equation of the polar of the circle is,
  2xx1+2yy1=112×4x+2×(1)y=118x2y=11  


    Question 6: Find the polar of the point (2,3) with respect to the circle    x2+y24x6y+5=0  
Solution: The equation of the polar of the circle is,    xx1+yy142(x+x1)62(y+y1)+5=02x+3y2(x2)3(y+3)+5=02x+3y2x+43y9+5=04x=0 x=0  


    Question 7: Find the pole of the straight line  2x+y+12=0   with respect to the circle  x2+y24x+3y1=0  
Solution: The given equations are, 
  2x+y+12=0 x2+y24x+3y1=0  
Let (x1,y1) be the pole so the line (7.1) must coincide with the point (x1,y1) whose equation is,  xx1+yy142(x+x1)+32(y+y1)1=0x(x12)+y(y1+32)2x1+32y11=0  
Since (7.1) and (7.3) are same we have,  x122=y1+321=2x132y1+112  
Taking  1st and 3rd part we get,  6x1+12=2x132y1+18x1+12+32y1+1=016x1+3y1+26=0  
Taking 2nd and 3rd part we get,  y1+32=2x132y1+11212y118=2x132y1+124y136=4x13y1+224y13y13624x1=04x1+27y1+38=0  
Solving  (7.4) and (7.5) we get,  x1=4937y1=178111  
The required pole is  (x1,y1)(4937,178111)  

    Question 8: If the polar of the point P with respect to the circle  x2+y22bx=0 touches the circle  x2+y2=a2 . Find the locus of P.
Solution: The given equations are  x2+y22bx=0 x2+y2=a2  
Let P(x1,y1) be the pole, then the polar of the circle  x2+y22bx=0 is  xx1+yy12b(x+x1)=0  
Since (8.1) touches (8.2) we have, 
  x1×0+y1×02b(0+x1)x12+y12=1 [since, length of the perpendicular is equal to the radius]2bx1=ax12+y124b2x12=a2(x12+y12)a2x12+a2y124b2x12=0(a24b2)x12+a2y12=0  

Find the equation of the direct common tangents to the circles  x2+y2=16 and x2+y2+6x78y=0  




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