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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 8th March 2019, 8:55 PM
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 Expert 9th March 2019, 12:22 PM
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Tags: circle tangent