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3 months ago

Find the point on the circle x^2+y^2 = 16900 which is closest to the interior point (30,40)

Mathematics

1 answer:

Leona [12.6K]3 months ago

5 0

Response-

(78,104) represents the point closest to the interior.

Explanation-

The equation defining the circle,

$\Rightarrow x^2+y^2 = 16900$

$\Rightarrow y^2 = 16900-x^2$

$\Rightarrow y = \sqrt{16900-x^2}$

Since the point lies on the circle, its coordinates must be,

$(x,\sqrt{16900-x^2})$

The distance "d" from the point to (30,40) can be calculated as,

$=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$=\sqrt{(x-30)^2+(\sqrt{16900-x^2}-40)^2}$

$=\sqrt{x^2+900-60x+16900-x^2+1600-80\sqrt{16900-x^2}}$

$=\sqrt{9400-60x-80\sqrt{16900-x^2}}$

Next, we need to determine the value of x for which d is minimized. The minimum distance occurs when $9400-60x-80\sqrt{16900-x^2}$ is at its lowest value.

Let’s set up the equation,

$\Rightarrow f(x)=9400-60x-80\sqrt{16900-x^2}$

$\Rightarrow f'(x)=-60+80\dfrac{x}{\sqrt{16900-x^2}}$

$\Rightarrow f''(x)=\dfrac{1352000}{\left(16900-x^2\right)\sqrt{16900-x^2}}$

We find the critical points,

$\Rightarrow f'(x)=0$

$\Rightarrow-60+80\dfrac{x}{\sqrt{16900-x^2}}=0$

$\Rightarrow 80\dfrac{x}{\sqrt{16900-x^2}}=60$

$\Rightarrow 80x=60\sqrt{16900-x^2}$

$\Rightarrow 80^2x^2=60^2(16900-x^2)$

$\Rightarrow 6400x^2=3600(16900-x^2)$

$\Rightarrow \dfrac{16}{9}x^2=16900-x^2$

$\Rightarrow \dfrac{25}{9}x^2=16900$

$\Rightarrow x=\sqrt{\dfrac{16900\times 9}{25}}=78$

$\Rightarrow x=78$

Then,

$\Rightarrow f''(78)=\dfrac{1352000}{\left(16900-78^2\right)\sqrt{16900-78^2}}=\dfrac{125}{104}=1.2$

Since f''(x) is positive, the function f(x) achieves its minimum at x=78

When x is set to 78, the corresponding y value will be

$\Rightarrow y = \sqrt{16900-x^2}=\sqrt{16900-78^2}=104$

This leads us to conclude that the closest point is (78,104)

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