MHB Finding the point on a circle closest to the given point

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The discussion focuses on finding the point on the circle defined by x² + y² = 16 that is closest to the point P(0,6). Participants analyze the distance formula and the constraints of the problem, noting that traditional differentiation methods may not yield the correct minimum distance due to the nature of the boundary conditions. There is a consensus that the book's answer of (+/-5, 3) is incorrect, and the approach of substituting y into the equation to find x does not produce valid results. Visualizing the problem through a plot of the circle and point is recommended for clarity. The conclusion emphasizes the importance of checking boundaries and understanding the geometric relationship between the point and the circle.
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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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