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  4. If real numbers x and y satisfy (x+5)2+(y-12)2=(14)2, then the minimum value of √x2+y2 is

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Q. If real numbers $x$ and $y$ satisfy $(x+5)^{2}+(y-12)^{2}=(14)^{2}$, then the minimum value of $\sqrt{x^{2}+y^{2}}$ is _____

Conic Sections

Solution:

Let $x+5=14 \cos \theta$ and $y-12=14 \sin \theta$
$\therefore x^{2}+y^{2}=(14 \cos \theta-5)^{2}+(14 \sin \theta+12)^{2}$
$=196+25+144+28(12 \sin \theta-5 \cos \theta)$
$=365+28(12 \sin \theta-5 \cos \theta)$
$\left.\therefore \sqrt{x^{2}+y^{2}}\right|_{\min }=\sqrt{365-28 \times 13}$
$=\sqrt{365-364}=1$