If-x-and-y-real-number-satisfy-x-5-2-y-12-2-196-then-the-minimum-value-of-x-2-y-2-is-

Question Number 96500 by bemath last updated on 02/Jun/20

If x a n d y r e a l n u m b e r s a t i s f y

{(x + 5)}^{2} + {(y - 12)}^{2} = 196, then

the minimum value of x^{2} + y^{2} i s

Answered by bobhans last updated on 02/Jun/20

Let x + 5 = 14 \cos θ, y - 12 = 14 \sin θ

x^{2} + y^{2} = 365 + 28 (12 \sin θ - 5 \cos θ)

= 365 + 364 \sin (θ - β) where \tan β = \frac{5}{12}

so x^{2} + y^{2} h a s t h e m i n i m u m v a l u e 1

w h e n θ = \frac{3 π}{2} + \arctan (\frac{5}{12}), i . e

\Rightarrow x = \frac{5}{13} & y = - \frac{12}{13}

Commented by bemath last updated on 02/Jun/20

thanks

Commented by 1549442205 last updated on 02/Jun/20

I add a question : Which is the greatest

value of the expression x^{2} + y^{2} ?

Commented by MJS last updated on 02/Jun/20

365 + 364 = 729

Commented by bobhans last updated on 02/Jun/20

yes ===

Commented by bobhans last updated on 02/Jun/20

when θ = \frac{π}{2} + \arctan \frac{5}{12}

Commented by 1549442205 last updated on 02/Jun/20

you are all right! it is as following :

we have x^{2} + 10 x + 25 + y^{2} - 24 y + 144 = 196

\Rightarrow x^{2} + y^{2} = 27 - 10 x + 24 y (1) . Applyin

{Bunhiacopxky}^{'} s inequality we get

{(- 10 x + 24 y)}^{2} ⩽ (10^{2} + 24^{2}) (x^{2} + y^{2})

= 676 (x^{2} + y^{2}) \Rightarrow∣ - 10 x + 24 y ∣⩽ 26 \sqrt{x^{2} + y^{2}} (2)

From (1), (2) we get x^{2} + y^{2} ⩽ 27

+ 26 \sqrt{x^{2} + y^{2}} \Leftrightarrow (\sqrt{x^{2} + y^{2}} + 1) (\sqrt{x^{2} + y^{2}} - 27) ⩽ 0

\Rightarrow \sqrt{x^{2} + y^{2}} ⩽ 27 \Rightarrow x^{2} + y^{2} ⩽ 729 .

The equality occurs if and only

if {\begin{cases} \frac{x}{- 10} = \frac{y}{24} \\ x^{2} + y^{2} = 729 \end{cases} \Leftrightarrow {\begin{cases} x = \frac{- 135}{13} \\ y = \frac{324}{13} \end{cases}

Answered by john santu last updated on 02/Jun/20

i have a method in short

if {(x - a)}^{2} + {(y - b)}^{2} = r^{2}

then x^{2} + y^{2} {\begin{cases} \max = {r + \sqrt{a^{2} + b^{2}}}^{2} \\ \min = {r - \sqrt{a^{2} + b^{2}}}^{2} \end{cases}

Commented by bobhans last updated on 02/Jun/20

waw ==== thanks

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