Question

The line 2x+3y=6 cuts x-axis at A and y-axis at B, the line Kx+8y =11 cuts x -axis at A' and y -axis at B´, then points A, B, A´, B´, will be con cyclic for

• A. Only one value of K
• B. Two values of K
• C. Three values of K
• D. If K = 12

Answer 1

Answer: (B), (D)

The correct options are
B Two values of K
D If K = 12

The Equation of conic through A, B, A´, B´ must be (2x+3y-6) (Kx+8y-11) + $\lambda xy=0$
If this represent a circle, coefficient of $x^2$ = coefficient of $y^2$ i.e. $2K=24\Rightarrow K=12$
But the points can also be concylic of( $\frac{11}{K}$ , 0) coincides with (3, 0), $\Rightarrow K=\frac{11}{3}$.