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Question Number 153189 by liberty last updated on 05/Sep/21

	Answered by EDWIN88 last updated on 05/Sep/21
	$eq of circle 16x^2 +16y^2 +48x−8y−43= with center point { ((x=−((48)/(32))=−(3/2))),((y=(8/(32))=(1/4))) :} with radius =(√((9/4)+(1/6)−(((−43)/(16))))) ⇒r=(√((36+1+43)/(16))) =((4(√5))/4)=(√5) let P((√5)cos θ−(3/2), (√5)sin θ+(1/4)) the distance point P to line r ⇒d=((∣8(√5)cos θ−12−4((√5)sin θ+(1/4))+73∣)/( (√(64+16)))) ⇒d=((∣8(√5) cos θ−4(√5)sin θ+60∣)/(4(√5))) minimum λ=8(√5) cos θ−4(√5)sin θ+60 is −(√(64×5+16×5)) +60=40 d_(min) =((40)/(4(√5))) = ((10)/( (√5)))=2(√5) when cos (θ+tan^(−1) ((1/2)))=−1 θ=π−tan^(−1) ((1/2))→ { ((cos θ=−tan^(−1) ((1/2))=−0.45)),((sin θ=tan^(−1) ((1/2))=0.45)) :}$
	$e q o f c i r c l e 16 x^{2} + 16 y^{2} + 48 x - 8 y - 43 =$ $w i t h c e n t e r p o i n t {\begin{cases} x = - \frac{48}{32} = - \frac{3}{2} \\ y = \frac{8}{32} = \frac{1}{4} \end{cases}$ $w i t h r a d i u s = \sqrt{\frac{9}{4} + \frac{1}{6} - (\frac{- 43}{16})}$ $\Rightarrow r = \sqrt{\frac{36 + 1 + 43}{16}} = \frac{4 \sqrt{5}}{4} = \sqrt{5}$ $l e t P (\sqrt{5} \cos θ - \frac{3}{2}, \sqrt{5} \sin θ + \frac{1}{4})$ $t h e d i s t a n c e p o i n t P t o l i n e r$ $\Rightarrow d = \frac{∣ 8 \sqrt{5} \cos θ - 12 - 4 (\sqrt{5} \sin θ + \frac{1}{4}) + 73 ∣}{\sqrt{64 + 16}}$ $\Rightarrow d = \frac{∣ 8 \sqrt{5} \cos θ - 4 \sqrt{5} \sin θ + 60 ∣}{4 \sqrt{5}}$ $m i n i m u m λ = 8 \sqrt{5} \cos θ - 4 \sqrt{5} \sin θ + 60$ $i s - \sqrt{64 \times 5 + 16 \times 5} + 60 = 40$ $d_{m i n} = \frac{40}{4 \sqrt{5}} = \frac{10}{\sqrt{5}} = 2 \sqrt{5}$ $w h e n \cos (θ + \tan^{- 1} (\frac{1}{2})) = - 1$ $θ = π - \tan^{- 1} (\frac{1}{2}) \to {\begin{cases} \cos θ = - \tan^{- 1} (\frac{1}{2}) = - 0.45 \\ \sin θ = \tan^{- 1} (\frac{1}{2}) = 0.45 \end{cases}$
	Commented by EDWIN88 last updated on 05/Sep/21

	$i n o t h e r w a y$ $d_{m i n} = d [(- \frac{3}{2}, \frac{1}{4}), 8 x - 4 y + 73 = 0] - r a d i u s$ $d [(- \frac{3}{2}, \frac{1}{4}), 8 x - 4 y + 73 = 0] =$ $\frac{∣ 8 (- \frac{3}{2}) - 4 (\frac{1}{4}) + 73 ∣}{4 \sqrt{5}} = \frac{60}{4 \sqrt{5}} = \frac{15}{\sqrt{5}} = 3 \sqrt{5}$ $d_{m i n} = 3 \sqrt{5} - \sqrt{5} = 2 \sqrt{5}$
	Answered by mr W last updated on 05/Sep/21

	$32 x + 32 {y y}^{'} + 48 - 8 y^{'} = 0$ $32 x + 32 y \times 2 + 48 - 8 \times 2 = 0$ $\Rightarrow x + 2 y + 1 = 0$ $16 {(- 2 y - 1)}^{2} + 16 y^{2} + 48 (- 2 y - 1) - 8 y - 43 = 0$ $16 y^{2} - 8 y - 15 = 0$ $y = \frac{5}{4}, - \frac{3}{4}$ $x = - \frac{7}{2}, \frac{1}{2}$ $d_{m i n} = \frac{∣ - 8 \times \frac{7}{2} - 4 \times \frac{5}{4} + 73 ∣}{\sqrt{8^{2} + 4^{2}}} = \frac{10}{\sqrt{5}}$ $f r o m p o i n t (- \frac{7}{2}, \frac{5}{4})$
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