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Question Number 35430 by ajfour last updated on 18/May/18

Commented by ajfour last updated on 19/May/18

$S o l u t i o n t o Q . 35422 (F i n d P Q)$
	Answered by ajfour last updated on 20/May/18

	$W e s h a l l o r i e n t t h e x y z c o o r d i n a t e$ $s y s t e m a s s h o w n w i t h A a s o r i g i n .$ $S o l i n e A P Q l i e s e n t i r e l y i n$ $x z p l a n e .$ $e q n . o f s p h e r e :$ ${(x - a \cos ϕ)}^{2} + {(y - a \sin ϕ)}^{2} + {(z - R)}^{2} = R^{2}$ $e q n . o f c i r c l e o f i n t e r s e c t i o n o f$ $s p h e r e w i t h x z p l a n e (y = 0) i s$ ${(x - a \cos ϕ)}^{2} + {(z - R)}^{2} = R^{2} - a^{2} \sin^{2} ϕ$ $e q u a t i o n o f A P Q i s$ $z = x \tan θ$ $F o r x c o o r d i n a t e s o f P a n d Q$ $(i n t e r s e c t i o n o f c i r c l e w i t h l i n e)$ ${(x - a \cos ϕ)}^{2} + {(x \tan θ - R)}^{2} = R^{2} - a^{2} \sin^{2} ϕ$ $x^{2} \sec^{2} θ - 2 x (s \cos ϕ + R \tan θ) + a^{2} = 0$ $P Q =∣ x_{Q} - x_{P} ∣ \sec θ$ $∣ x_{Q} - x_{P} ∣= \sqrt{{(x_{Q} + x_{P})}^{2} - 4 x_{B} x_{P}}$ $\Rightarrow P Q = (\sec θ) \sqrt{{[\frac{2 (a \cos ϕ + R \tan θ)}{\sec^{2} θ}]}^{2} - \frac{4 a^{2}}{\sec^{2} θ}}$ $P Q = 2 \sqrt{{(a \cos ϕ + R \tan θ)}^{2} \cos^{2} θ - a^{2}} .$
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