The points $(4, - 2) (3, b)$ are conjugate with respect to the circle $x^{2} + y^{2} = 24$ , if $b =$

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Solution

The correct option is A $- 6$
Let. $A (4, - 2)$ and $B (3, b)$ are conjugate w.r.t $x^{2} + y^{2} = 24$
co, by property of conjugate points $A$ lies polar of $B$
chord of contact to $x^{2} + y^{2} = 24$ from $(3, b)$ is
$3 x + y b = 24$
So, $A (4, - 2)$ lie on $3 x + y b = 24$
So, $12 - 2 b = 2$
$2 b = - 12$
$\Rightarrow b = - 6$

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The points (4,−2)(3,b) are conjugate with respect to the circle x2+y2=24, if b=

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