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Question Number 168609 by infinityaction last updated on 14/Apr/22

Commented by infinityaction last updated on 14/Apr/22

$c o o r d i n a t e o f p o i n t (a, b)$
	Answered by mr W last updated on 14/Apr/22

	$r = \sqrt{{(12 - 6)}^{2} + {(- 6 - 4)}^{2}} = 2 \sqrt{34}$ $\sin α = \frac{4 - (- 6)}{2 \sqrt{34}} = \frac{5}{\sqrt{34}}$ $\cos α = \frac{6 - 12}{2 \sqrt{34}} = - \frac{3}{\sqrt{34}}$ $a = 12 + 2 \sqrt{34} \cos (α + \frac{π}{3})$ $= 12 + 2 \sqrt{34} (- \frac{3}{\sqrt{34}} \times \frac{1}{2} - \frac{5}{\sqrt{34}} \times \frac{\sqrt{3}}{2})$ $= 9 - 5 \sqrt{3}$ $b = - 6 + 2 \sqrt{34} \sin (α + \frac{π}{3})$ $= - 6 + 2 \sqrt{34} (\frac{5}{\sqrt{34}} \times \frac{1}{2} - \frac{3}{\sqrt{34}} \times \frac{\sqrt{3}}{2})$ $= - 1 - 3 \sqrt{3}$
	Commented by infinityaction last updated on 14/Apr/22

	$w h e r e i s a α i n t h i s d i a g r a m$
	Commented by mr W last updated on 14/Apr/22

	Commented by infinityaction last updated on 14/Apr/22

	$t h a n k y o u s i r$
	Answered by cortano1 last updated on 14/Apr/22
	$let ℓ is line passes througt point B(((6+12)/2),((4−6)/2))=B(9,−1) with gradient=−(1/((((4+6)/(6−12))))) =(3/5) then ℓ ≡ 3x−5y=32 equation of circle (x−12)^2 +(y+6)^2 =136 then the line ℓ cut the circle ⇒(((32+5y)/3)−12)^2 +(y+6)^2 =136 ⇒(((5y−4)/3))^2 +(y+6)^2 =136 ⇒ { ((y_1 = −3(√3)−1 = b)),((x_1 = −5(√3)+9 = a)) :} ∴ P(−5(√3)+9, −3(√3)−1)$
	$l e t ℓ i s l i n e p a s s e s t h r o u g t p o i n t$ $B (\frac{6 + 12}{2}, \frac{4 - 6}{2}) = B (9, - 1)$ $w i t h g r a d i e n t = - \frac{1}{(\frac{4 + 6}{6 - 12})} = \frac{3}{5}$ $t h e n ℓ \equiv 3 x - 5 y = 32$ $e q u a t i o n o f c i r c l e$ ${(x - 12)}^{2} + {(y + 6)}^{2} = 136$ $t h e n t h e l i n e ℓ c u t t h e c i r c l e$ $\Rightarrow {(\frac{32 + 5 y}{3} - 12)}^{2} + {(y + 6)}^{2} = 136$ $\Rightarrow {(\frac{5 y - 4}{3})}^{2} + {(y + 6)}^{2} = 136$ $\Rightarrow {\begin{cases} y_{1} = - 3 \sqrt{3} - 1 = b \\ x_{1} = - 5 \sqrt{3} + 9 = a \end{cases}$ $∴ P (- 5 \sqrt{3} + 9, - 3 \sqrt{3} - 1)$
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