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Question Number 102078 by Jamshidbek2311 last updated on 06/Jul/20

Answered by mr W last updated on 06/Jul/20

$$\mathit{M}\mathit{E}\mathit{T}\mathit{H}\mathit{O}\mathit{D}\mathit{I}$$$$x^2+y^{2}isthesquareofthedistance$$$$frompoint(x,y)totheorigin(0,0).$$$$$$$${(x+5)}^2+{(y-12)}^2={14}^{2}isthecircle$$$$withradius14andcenterat(-5,12).$$$$distancefrom(-5,12)totheorigin$$$$is\sqrt{{(-5)}^2+{12}^2}=13,theminimum$$$$distancefromapointonthecircleto$$$$theoriginisthen14-13=1andthe$$$$maximumdistanceis14+13=27.$$$$\Rightarrow {(x^2+y^2)}_{min}=1^2=1$$$$\Rightarrow {(x^2+y^2)}_{max}={27}^2=729$$

Answered by mr W last updated on 06/Jul/20

$$\mathit{M}\mathit{E}\mathit{T}\mathit{H}\mathit{O}\mathit{D}\mathit{I}\mathit{I}$$$$k=x^2+y^2$$$$F=x^2+y^2+\lambda [{(x+5)}^2+{(y-12)}^2-{14}^2]$$$$\frac{\partial F}{\partial x}=2x+2\lambda (x+5)=0\Rightarrow x=-\frac{5\lambda }{\lambda +1}$$$$\frac{\partial F}{\partial y}=2y+2\lambda (y-12)=0\Rightarrow y=\frac{12\lambda }{\lambda +1}$$$${(-\frac{5\lambda }{\lambda +1}+5)}^2+{(\frac{12\lambda }{\lambda +1}-12)}^2-{14}^2=0$$$$5^2{(1-\frac{\lambda }{\lambda +1})}^2+{12}^2{(1-\frac{\lambda }{\lambda +1})}^2-{14}^2=0$$$${13}^2{(1-\frac{\lambda }{\lambda +1})}^2={14}^2$$$$\Rightarrow 1-\frac{\lambda }{\lambda +1}=\pm \frac{14}{13}$$$$\Rightarrow \lambda +1=\pm \frac{13}{14}$$$$\Rightarrow \lambda =-\frac{1}{14}or-\frac{27}{14}$$$$with\lambda =-\frac{1}{14}:$$$$x=-5\times \frac{-\frac{1}{14}}{1-\frac{1}{14}}=\frac{5}{13}$$$$y=12\times \frac{-\frac{1}{14}}{1-\frac{1}{14}}=-\frac{12}{13}$$$$\Rightarrow k=x^2+y^2={\left(\frac{5}{13}\right)}^2+{(-\frac{12}{13})}^2=1=min.$$$$with\lambda =-\frac{27}{14}:$$$$x=-5\times \frac{-\frac{27}{14}}{1-\frac{27}{14}}=-5\times \frac{27}{13}$$$$y=12\times \frac{-\frac{27}{14}}{1-\frac{27}{14}}=12\times \frac{27}{13}$$$$\Rightarrow k=x^2+y^2={(-5\times \frac{27}{13})}^2+{\left(12\times \frac{27}{13}\right)}^2=729=max.$$

Answered by mr W last updated on 06/Jul/20

$$\mathit{M}\mathit{E}\mathit{T}\mathit{H}\mathit{O}\mathit{D}\mathit{I}\mathit{I}\mathit{I}$$$$k=x^2+y^2=d^2$$$$letx=d\cos \theta ,y=d\sin \theta$$$${(d\cos \theta +5)}^2+{(d\sin \theta -12)}^2={14}^2$$$$d^2+2d(5\cos \theta -12\sin \theta )-27=0$$$$d^2+26d(\sin \alpha \cos \theta -\cos \alpha \sin \theta )-27=0$$$$d^2+26d\sin (\alpha -\theta )-27=0$$$$d=-13\sin (\alpha -\theta )+\sqrt{169{\sin }^2(\alpha -\theta )+27}$$$$d_{max}=13+\sqrt{169+27}=13+14=27$$$$d_{min}=-13+\sqrt{169+27}=-13+14=1$$$$k=x^2+y^2=d^2={27}^2=729=max.$$$$k=x^2+y^2=d^2=1^2=1=min.$$

Answered by mr W last updated on 06/Jul/20

$$\mathit{M}\mathit{E}\mathit{T}\mathit{H}\mathit{O}\mathit{D}\mathit{I}\mathit{V}$$$$x=-5+14\cos \theta$$$$y=12+14\sin \theta$$$$x^2+y^2=5^2+{12}^2+{14}^2+28(-5\cos \theta +12\sin \theta )$$$$=365+28\times 13(-\sin \alpha \cos \theta +\cos \alpha \sin \theta )$$$$=365+28\times 13\sin (\theta -\alpha )$$$$max.=365+28\times 13=729$$$$min.=365-28\times 13=1$$

Answered by john santu last updated on 07/Jul/20

$$methodJS$$$$x^2+y_{min}^2=[r-\sqrt{x_c^2+y_c^2}]{}^2$$$$x^2+y_{max}^2={[r+\sqrt{x_c^2+y_c^2}]}^2$$

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