Question

If ax²+4xy+2y²+x+y+5=0 and ax²+6xy+5y²+2x+3y+8=0 intersect at four concyclic points, then the value of a is?

Answer 1

Answer:

ax2+4xy+2y2+x+y+5+ K(ax2+6xy+5y2+2x+3y+8) =0

If this curve represent a circle then coiffiecient of xy should be 0 and coiffiecient of x2 and y2 should be equal.

6K=-4,

K=-2/3

and a/3= 2-10/3

a/3 =-4/3

a=-4

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