lf two conics whose equations refer to rect- angular axes are ax2+2hxy+by2+2gx+2fy+c=0 and a′x2+2h′xy+b′y2+2g′x+2fy+c′=0 intersect in four concyclic points, then
A
h′(a−b)=h(a′−b′)
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B
h(a−b)=h′(a′−b′)
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C
h(a′−b)=h′(a−b′)
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D
h(a−b′)=h′(a′−b)
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Solution
The correct option is Ah′(a−b)=h(a′−b′) S1:ax2+2hxy+by2+2gx+2fy+c=0 S2:a′x2+2h′xy+b′y2+2g′x+2fy+c′=0 four points intersects on the curve: S1+kS2=0 ax2+2hxy+by2+2gx+2fy+c+k(a′x2+2h′xy+b′y2+2g′x+2fy+c′)=0 Since, they are con-cyclic points Therefore, coefficient of x2= coefficient of y2 and coefficient of xy=0 a+ka′=b+kb′ 2h+2kh′=0 Eliminating k, we get h′(a−b)=h(a′−b′)