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Q) Prove that the tangent to the circle x^2+y^2=5 at the point (1,-2) also touches the circle x^2+y^2-8x+6y+20=0 and find its point of contact.
Asked by Anish | 08 Mar, 2019, 08:55: PM
Expert Answer
Tangent to circle x2 + y2 = 5 at a point (x1 , y1 ) on circle is xx1 + yy1 = 5
we need a tangent at a point(1, -2) on circle. hence tangent is x - 2y = 5 ............................(1)
If eqn.(1) is also tangent to circle x2 +y2 -8x +6y +20 = 0, then we substitute y = (x-5)/2 in the circle eqn. and
get the point of contact of this tangent
x2 +[ (x-5)2 /4 ] -8x +6 [ (x-5)/2 ] + 20 = 0
after simplification of the above eqn., we get, x2 -6x +9 = 0 or (x-3) = 0 or x = 3
By substituting x =3 in eqn.(1), we get y = -1
Hence point of contact of the tangent given by eqn.(1) to the circle x2 +y2 -8x +6y +20 = 0, is (3, -1)
Verification of eqn.(1) is tangent to circle x2 +y2 -8x +6y +20 = 0 is done by substituting coordinates of point of contact
in tangent equation xx1 + yy1-4(x+x1) +3(y+y1) +20 = 0
x(3) + y(-1) -4(x+3) +3(y-1) +20 = -x +2y + 5 = 0 or x-2y = 5 which is same as eqn.(1)
Answered by Thiyagarajan K | 09 Mar, 2019, 12:22: PM
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