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

Commented by infinityaction last updated on 14/Apr/22

coordinate of point (a,b)

$${coordinate}\:{of}\:{point}\:\:\left({a},{b}\right)\: \\ $$

Answered by mr W last updated on 14/Apr/22

r=(√((12−6)^2 +(−6−4)^2 ))=2(√(34)) sin α=((4−(−6))/(2(√(34))))=(5/( (√(34)))) cos α=((6−12)/(2(√(34))))=−(3/( (√(34)))) a=12+2(√(34)) cos (α+(π/3)) =12+2(√(34))(−(3/( (√(34))))×(1/2)−(5/( (√(34))))×((√3)/2)) =9−5(√3) b=−6+2(√(34)) sin (α+(π/3)) =−6+2(√(34)) ((5/( (√(34))))×(1/2)−(3/( (√(34))))×((√3)/2)) =−1−3(√3)

$${r}=\sqrt{\left(\mathrm{12}−\mathrm{6}\right)^{\mathrm{2}} +\left(−\mathrm{6}−\mathrm{4}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{34}} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{4}−\left(−\mathrm{6}\right)}{\mathrm{2}\sqrt{\mathrm{34}}}=\frac{\mathrm{5}}{\:\sqrt{\mathrm{34}}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{6}−\mathrm{12}}{\mathrm{2}\sqrt{\mathrm{34}}}=−\frac{\mathrm{3}}{\:\sqrt{\mathrm{34}}} \\ $$$${a}=\mathrm{12}+\mathrm{2}\sqrt{\mathrm{34}}\:\mathrm{cos}\:\left(\alpha+\frac{\pi}{\mathrm{3}}\right) \\ $$$$\:\:=\mathrm{12}+\mathrm{2}\sqrt{\mathrm{34}}\left(−\frac{\mathrm{3}}{\:\sqrt{\mathrm{34}}}×\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{5}}{\:\sqrt{\mathrm{34}}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\:\:=\mathrm{9}−\mathrm{5}\sqrt{\mathrm{3}} \\ $$$${b}=−\mathrm{6}+\mathrm{2}\sqrt{\mathrm{34}}\:\mathrm{sin}\:\left(\alpha+\frac{\pi}{\mathrm{3}}\right) \\ $$$$\:\:=−\mathrm{6}+\mathrm{2}\sqrt{\mathrm{34}}\:\left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{34}}}×\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{34}}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\:\:=−\mathrm{1}−\mathrm{3}\sqrt{\mathrm{3}} \\ $$

Commented by infinityaction last updated on 14/Apr/22

where is a α in this diagram

$${where}\:{is}\:{a}\:\alpha\:{in}\:{this}\:{diagram} \\ $$

Commented by mr W last updated on 14/Apr/22

Commented by infinityaction last updated on 14/Apr/22

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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)

$${let}\:\ell\:{is}\:{line}\:{passes}\:{througt}\:{point} \\ $$$$\:{B}\left(\frac{\mathrm{6}+\mathrm{12}}{\mathrm{2}},\frac{\mathrm{4}−\mathrm{6}}{\mathrm{2}}\right)={B}\left(\mathrm{9},−\mathrm{1}\right) \\ $$$$\:{with}\:{gradient}=−\frac{\mathrm{1}}{\left(\frac{\mathrm{4}+\mathrm{6}}{\mathrm{6}−\mathrm{12}}\right)}\:=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\:{then}\:\ell\:\equiv\:\mathrm{3}{x}−\mathrm{5}{y}=\mathrm{32} \\ $$$${equation}\:{of}\:{circle}\: \\ $$$$\:\left({x}−\mathrm{12}\right)^{\mathrm{2}} +\left({y}+\mathrm{6}\right)^{\mathrm{2}} =\mathrm{136} \\ $$$$\:{then}\:{the}\:{line}\:\ell\:{cut}\:{the}\:{circle} \\ $$$$\Rightarrow\left(\frac{\mathrm{32}+\mathrm{5}{y}}{\mathrm{3}}−\mathrm{12}\right)^{\mathrm{2}} +\left({y}+\mathrm{6}\right)^{\mathrm{2}} =\mathrm{136} \\ $$$$\Rightarrow\left(\frac{\mathrm{5}{y}−\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} +\left({y}+\mathrm{6}\right)^{\mathrm{2}} =\mathrm{136}\: \\ $$$$\:\Rightarrow\begin{cases}{{y}_{\mathrm{1}} =\:−\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{1}\:=\:{b}}\\{{x}_{\mathrm{1}} =\:−\mathrm{5}\sqrt{\mathrm{3}}+\mathrm{9}\:=\:{a}}\end{cases} \\ $$$$\:\therefore\:{P}\left(−\mathrm{5}\sqrt{\mathrm{3}}+\mathrm{9},\:−\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$

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