on-circle-x-2-y-2-25-find-the-point-closest-to-1-1-

Question Number 132407 by bramlexs22 last updated on 14/Feb/21

on circle x^{2} + y^{2} = 25, find the point

closest to (1, 1) .

Answered by EDWIN88 last updated on 14/Feb/21

let P (x, y) be a point on a circle

let x be a distance of plint (1, 1) to (x, y)

d^{2} = {(x - 1)}^{2} + {(y - 1)}^{2}

d^{2} = {(x - 1)}^{2} + 25 - x^{2} - 2 \sqrt{25 - x^{2}} + 1

d^{2} = x^{2} - 2 x + 1 + 25 - x^{2} - 2 \sqrt{25 - x^{2}} + 1

d^{2} = 27 - 2 x - 2 \sqrt{25 - x^{2}}

differentiating w . r . t x

{(d^{2})}^{'} = - 2 - 2 (\frac{- x}{\sqrt{25 - x^{2}}}) = 0

\frac{x}{\sqrt{25 - x^{2}}} = 1 \Leftrightarrow {2 x}^{2} = 25 \Rightarrow x = \pm \frac{5}{\sqrt{2}}

and y = \pm \sqrt{25 - \frac{25}{2}} = \pm \frac{5}{\sqrt{2}} the point

on circle such that closest to (1, 1) is (\frac{5}{\sqrt{2}}, \frac{5}{\sqrt{2}})

Commented by EDWIN88 last updated on 14/Feb/21

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