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Question

Prove that the equation x2+y22x2ay8=0 represents a family of circles passing through two fixed points, say P and Q. Choose the member of the family tangents to which at P and Q intersect on the line x+2y+5=0.

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Solution

The given circle is
(x2+y22x8)2ay=0.....(1)
It is of the form S+λP=0 and hence represents a family of circles passing through the points of intersection S=0 and P=0.
i.e. x2+y22x8=0 and y=0.
P and Q are (2,0) and (4,0).
Equation of PQ is y=0.....(2)
If tangents at P and Q intersect at R(h,k) then PQ is chord of contact of (h,k) whose equation is x.h+y.k(x+h)a(y+k)8=0
or x(h1)+y(ka)(h+ak+8)=0.....(3)
Comparing (3) and (2),h1=0,h+ak+8=0,ka= any constant,
h=1. If (h,k) lies on x+2y+5=0
h+2k+5=0k=3. Putting for (h,k) in h+ak+8=0, we get a=3.
Hence the required member of the family of circles is x2+y22x6y8=0. by (1).
923507_1007644_ans_a9b3c2ab5416484fb370a74b122f8d69.png

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