Two conics `a_1x^2+2h_1xy + b_1y^2 = c_1, a_2x^2 + 2h_2xy+b_2y^2 = c_2` intersect in 4 concyclic points. Then

Options

A. `(a_(1)-b_(1)h_(2)=(a_(2)-b_(2))h_(1)`
B. `(A_(1)-b_(1))h_(1)=(a_(2)-b_(2))h_(2)`
C. `(a_(1)+b_(1))h_(2)=(a_(2)+b_(2))h_(1)`
D. `(a_(1)+b_(1))h_(1)=(a_(2)+b_(2))h_(2)`

Answer

A

Explanation

The equation of a conic passing through the intersection of the given conics is
`(a_(1)x^(2)+2h_(1)xy+b_(1)y^(2)-c_(1))+lambda(a_(2)x^(2)+2h_(2)xy+b_(2)y^(2)-c_(2))=0`
This equation will represent a circle, if
Coeff. of `x^(2)=`Coeff. of `y^(2)`and Coeff. of `xy=0`
`rArr (a_(1)+lambda a_(2))=(b_(1)+lambda b_(2))` and `2h_(1)+2lambda h_(2)=0`
Eliminating `lambda` from these two equations, we get
`(a_(1)-b_(1))=(h_(1))/(h_(2))(a_(2)-b_(2))rArr (a_(1)-b_(1))h_(2)=(a_(2)-b_(2))h_(1)`