Finding the point on a circle closest to the given point

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SUMMARY

The discussion centers on finding the point on the circle defined by the equation x² + y² = 16 that is closest to the point P(0, 6). Participants clarify that the correct approach involves substituting y = √(16 - x²) into the distance formula f(x, y) = (x - 0)² + (y - 6)². The conclusion reached is that the book's answer of (+/-5, 3) is incorrect, as the minimum distance cannot be determined solely through differentiation, indicating a boundary minimum scenario. Visualizing the problem by plotting the circle and the point is recommended for clarity.

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leprofece
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Find the curve coordinates of the point nearest to P in the circle
x2 + y2 = 16 P(0,6)

as the former ( see a gift )

x2 + (y-6)2 = 16 (1)

solving for y = y2 = 16- x2

introducing en 1 x2 +(16-x-6)2 = 16
x2 +100-20x + x2 = 16

derivating
4x -20
and x = 5

y = sqrt ( 16-25) and i Got stuck

The book ANSWER Is (+/-5, 3)
 
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I have retitled both threads to give them titles that describe the problems being discussed.

I would define a point on the circle as:

$$(x,y)$$

The square of the distance between this point and the given point is:

$$f(x,y)=(x-0)^2+(y-6)^2=x^2+(y-6)^2$$

Now, our constraint may be written as:

$$x^2=16-y^2$$

and so we have:

$$f(y)=16-y^2+y^2-12y+36=4(13-3y)$$

Differentiation will not help here. So check your boundaries...

The answer you cite from your book makes no sense.
 
leprofece said:
solving for y = y2 = 16- x2

You cannot arbitrarily set $y$ equal to $y^2$.
If you continue your argument with $y = \sqrt{16- x^2}$, you will find the answer.
However, the answer is different from the book answer, which is wrong, as Mark already noted.

In this particular problem this approach works when substituting y and solving for x, but not when substituting x and solving for y.
In a sense it still works when substituting x, but the minimum distance cannot be found with a derivative. It is a so called boundary minimum - for shorter distances there is no solution.

I suggest to draw the curve and the point.
You will be able to see immediately what the shortest distance is.
 
MarkFL said:
I have retitled both threads to give them titles that describe the problems being discussed.

I would define a point on the circle as:

$$(x,y)$$

The square of the distance between this point and the given point is:

$$f(x,y)=(x-0)^2+(y-6)^2=x^2+(y-6)^2$$

Now, our constraint may be written as:

$$x^2=16-y^2$$

and so we have:

$$f(y)=16-y^2+y^2-12y+36=4(13-3y)$$

Differentiation will not help here. So check your boundaries...

The answer you cite from your book makes no sense.

According to last part of the answer
y = 13/3
If i don't misunderstand
ok But $$x^2=16-y^2$$
then 16 -(13/3)
16 -18,77
So x does not have answer
 
Did you take I like Serena's advice and make a plot of the point and the circle? This was what I initially did and how I knew immediately the answer you gave is not correct.
 

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