# Thread: Smallest distance from origin to point on a given circle

1. ## Smallest distance from origin to point on a given circle

Hi all ---

I'm posting this question on geometry here because I'm not sure if you need calculus. But even if you don't, I'd like to know if I can use calculus to do the question.

Question with solution:

Now, I understand perfectly how to complete the square. So when I did this problem, I graphed it and saw that the point on the circle closest to $(0, 0)$ had to be in the 1st quadrant. This is because I can see that the origin is in the upper-right "quarter" of the circle, if I separate the circle into equal 4s.

But I'm not sure how to find the specific point in the 1st quadrant? The solution somehow knows that $(3,4)$ is the answer, but it doesn't explain how it got it?

Also, I know from calculus that ---
$\text{distance} = D = \sqrt{(x - a)^2 + (y - b)^2}$ where $x, y$ must be on the circle.

But I can maximize the square instead ---
$d = D^2 = (x - a)^2 + (y - b)^2}$ where $x, y$ must be on the circle.

But this doesn't seem to help, because it looks too complicated to sub in the equation of the circle into $d$?

Thanks ---

2. ## Re: Smallest distance from origin to point on a given circle

The line that determined by the origin and the center contains the point on the circle closest to the origin.

3. ## Re: Smallest distance from origin to point on a given circle

Hi Plato ---

But to be honest, I sort of see what you say is true. But at the same time, how can I be sure that some other line - passing through (0,0) and some point very near the centre of the circle - doesn't contain the point on the circle closest to (0,0)?

I've tried to look this topic up here but couldn't find anything - Is there a way to prove what you wrote?

Thanks again ---

4. ## Re: Smallest distance from origin to point on a given circle

Originally Posted by mathminor827
Also, I know from calculus that ---
$\text{distance} = D = \sqrt{(x - a)^2 + (y - b)^2}$ where $x, y$ must be on the circle.
But I can maximize the square instead ---
$d = D^2 = (x - a)^2 + (y - b)^2}$ where $x, y$ must be on the circle. But this doesn't seem to help, because it looks too complicated to sub in the equation of the circle into $d$?
No matter how complicated that is, it has to be done if you require a rigorous proof.
Originally Posted by mathminor827
How can I be sure that some other line - passing through (0,0) and some point very near the centre of the circle - doesn't contain the point on the circle closest to (0,0)? Is there a way to prove what you wrote?
If the circle is $(x-a)^2+(y-b)^2=r^2$ and $a\ne 0$ then the equation of the line determined by $(a,b)~\&~(0,0)$ is $y=\frac{b}{a}x$.
The slope of the tangent to the circle is $y'=-\frac{x-a}{y-b}$.
If you find where $y=\frac{b}{a}x$ intersects the circle you will find that line and the tangents the are perpendicular at the whose two points.
You will learn in vector geometry that one of points is the closest point on the circle to $(0,0)$ and the other is at the greatest distance.