MATH -127 (Coordinate Geometry)
Chapter -3The Circle
• In this figure, "O" is the pole and "OP" is the polar ( can be tangent to the circle)
• Length and midpoint of the chord intercepted by
x2+y2=a2
on the line
y=mx+c
Midpoint
(−mc1+m2,c1+m2)
AB=2AC=2√{a2(1+m2)−c21+m2}OC→ y=−1mx ∴ x+my=0
Since, the equation of AB is, y=mx+c , the equation of the perpendicular OC is y=−1mx . Solving (2) and (3) we will get the point C. The length of OA is a (since radius). Thus we can find A.
The equation of the polar with respect to the circle is x2+y2=a2 is xx1+yy1=a2 at (x1,y1)
Question 1: Find the pole of the straight line 2x−y=6 with respect to the circle 5x2+5y2=9 ( ---------- )
Solution: Given equations are,
2x−y=6 5x2+5y2=9
Let P(x1,y1 be the pole, then the equation of the pole with respect to the circle is given by
xx1+yy1=a2⟹5xx1+5yy1=9
Since (1.2) & (1.3) is identical, we can write
5x12=5y1−1=96
∴ the coordinates of the pole are
(35,−310)
♦ Combined equation of the pair of tangents:
x2+y2=a2⟹x2+y2−a2=0And,x2+y2+2gx+2fy+c=0⟹S=0
for Equation (5),SS1=T2⟹(x2+y2−a2)(x12+y12−a2)=(ss1+yy1−a2)2for Equation (6),SS1=T2⟹(x2+y2+2gx+2fy+c).(x12+y12+2gx1+2fy1+c) ={xx1+yy1+g(x+x1)+f(y+y1)+c}
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Question 2: Find the equation of the pair of the tangent drawn from (0,0) to
x2+y2+2gx+2fy+c=0
Solution: The equation of the pair of tangents drawn from (0,0) to
x2+y2+2gx+2fy+c=0
SS1=T2⟹(x2+y2+2gx+2fy+c).c={x×0+y×0+g(x+0)+f(y+0)+c}2⟹c.(x2+y2+2gx+2fy+c)=(gx+fy+c)2⟹cx2+cy2+2cgx+2cfy+c2=g2x2+f2g2+c2+2gfxy+2fy+2gxc⟹cx2+cy2−g2x2−f2y2−2gxfy=0⟹c(x2+y2)−(gx+fy)2=0⟹C.(x2+y2)=(gx+fy)2
Question 3: Prove that, if the polar of a point "P" with respect to the circle
x2+y2=37
touchhes the circle
(x−3)2+(y+2)2=2
. The locas of "P" must be conic.
Solution: From (1) we get,
xx1+yy1=37|3x1−2y1−37|√x12+y12⟹(3x1−2y1−37)2=25(x12+y12)⟹25x12+25y12+(−3x1+2y1+37)2=0⟹25x12+25y12+9x12+4y12+1369−12x1y1+148y1−111x1=0⟹34x12+29y12−111x1+148y1+1369=0
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