Finding the point on a circle closest to the given point

[leprofece]

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)

 

[MarkFL]

I have retitled both threads to give them titles that describe the problems being discussed.

I would define a point on the circle as:

\(\displaystyle (x,y)\)

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

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

Now, our constraint may be written as:

\(\displaystyle x^2=16-y^2\)

and so we have:

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

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

The answer you cite from your book makes no sense.

 

[I like Serena]

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.

 

[leprofece]

I have retitled both threads to give them titles that describe the problems being discussed.

I would define a point on the circle as:

\(\displaystyle (x,y)\)

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

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

Now, our constraint may be written as:

\(\displaystyle x^2=16-y^2\)

and so we have:

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

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 \(\displaystyle x^2=16-y^2\)
then 16 -(13/3)
16 -18,77
So x does not have answer

 

[MarkFL]

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.