• 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'.
locus is the line that is made of a 'set of points'.
Question 1: Find the pole of the line
2x−y=6
with respect to the circle
5x2+5y2=9
Solution: given that,
2x−y=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=5y1−2=96⟹x1=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]2⟹c(x2+y2)+c(2gx+2fy+c)=(gx+fy+c)2+(gx+fy)2+2(gx+fy)c+c2⟹c(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
(x−3)2+(y+2)2=25
then the locus of p must be a cone.
Solution: Given circles are
x2+y2=37 (x−3)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,
3x1−2y1−37√x12+y12
Since length of the perpendicular from (3,−2) to (3.1) is equal to the radius of the circle. so that,
(3x1−2y1−37)2=25(x12+y12)⟹16x12+21y12+12xy+222x1−148y1−1369=0
∴ the required locus is,
16x12+21y12+12xy+222x1−148y1−1369=0
Question 4: Find the pole of the straight line
9x+y=28
with respect to the circle
2x2+2y2−3x+5y−7=0
Solution: The given equations are
9x+y=28 2x2+2y2−3x+5y−7=0
If (x1,y1) be the required point the line (4.1) must coincide with the polar of (x1,y1) whose equation is,
2xx1+2yy1−32(x1+x)+52(y1+y)−7=0⟹x(4x1−3)+y(4y1+5)−3x1+4y1−14=0
Since (4.1) and (4.3) are the same, we have
4x1−39=4y1+51=−3x1+5y1−14−28
Hence,
x1=9y1+143x1−117y1=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=11⟹2×4x+2×(−1)y=11⟹8x−2y=11
Question 6: Find the polar of the point (−2,3) with respect to the circle
x2+y2−4x−6y+5=0
Solution: The equation of the polar of the circle is,
xx1+yy1−42(x+x1)−62(y+y1)+5=0⟹−2x+3y−2(x−2)−3(y+3)+5=0⟹−2x+3y−2x+4−3y−9+5=0⟹−4x=0∴ x=0
Question 7: Find the pole of the straight line
2x+y+12=0
with respect to the circle
x2+y2−4x+3y−1=0
Solution: The given equations are,
2x+y+12=0 x2+y2−4x+3y−1=0
Let (x1,y1) be the pole so the line (7.1) must coincide with the point (x1,y1) whose equation is,
xx1+yy1−42(x+x1)+32(y+y1)−1=0⟹x(x1−2)+y(y1+32)−2x1+32y1−1=0
Since (7.1) and (7.3) are same we have,
x1−22=y1+321=2x1−32y1+1−12
Taking 1st and 3rd part we get,
−6x1+12=2x1−32y1+1⟹−8x1+12+32y1+1=0⟹−16x1+3y1+26=0
Taking 2nd and 3rd part we get,
y1+32=2x1−32y1+1−12⟹−12y1−18=2x1−32y1+1⟹−24y1−36=4x1−3y1+2⟹−24y1−3y1−36−2−4x1=0⟹4x1+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+y2−2bx=0
touches the circle
x2+y2=a2
. Find the locus of P.
Solution: The given equations are
x2+y2−2bx=0 x2+y2=a2
Let P(x1,y1) be the pole, then the polar of the circle
x2+y2−2bx=0
is
xx1+yy1−2b(x+x1)=0
Since (8.1) touches (8.2) we have,
x1×0+y1×0−2b(0+x1)√x12+y12=1 [since, length of the perpendicular is equal to the radius]−2bx1=a√x12+y124b2x12=a2(x12+y12)a2x12+a2y12−4b2x12=0(a2−4b2)x12+a2y12=0
Find the equation of the direct common tangents to the circles
x2+y2=16 and x2+y2+6x−78y=0
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