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A circle has centre $(2,4)$. Prove that if $(0,3)$ is not inside the circle, then $(3,1)$ is not inside the circle. I just want to know if my method would be correct.

The method I used is as follows:

First I found the distance from the centre to $(0,3)$. Then I assumed that the radius of the circle is less than the distance found. Then I found the distance from the centre to $(3,1)$ and showed that this distance is greater than the radius. Hence if $(0,3)$ does not lie inside the circle then $(3,1)$ does not lie inside the circle.

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  • $\begingroup$ The idea is fine, there is currently a typo, you mean you found the distance from the centre to $(0,3)$. $\endgroup$ Commented Oct 18, 2014 at 8:01
  • $\begingroup$ ^Yup there was a typo. Thanks a million. :) $\endgroup$
    – Alexander.
    Commented Oct 18, 2014 at 8:31

2 Answers 2

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Let $r$ = the circle's radius, then : $(x-2)^2 +(y-4)^2 = r^2$, then since $A = (0,3) \notin \text{int(C)}$, we must have that $r < AO$ with $O = (2,4)$. But $AO = \sqrt{(2-0)^2 + (4-3)^2} = \sqrt{5}$, thus $r < \sqrt{5}$. And let $D = (3,1)$, then $DO = \sqrt{(3-2)^2 + (4-1)^2} = \sqrt{10}$. Since $\sqrt{5} < \sqrt{10}$, it follows that $r < \sqrt{10} = DO$, thus $D = (3,1) \notin \text{int(C)}$, whereas $\text{int(C)}$ is the interior of the circle $\text{C}$.

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  • $\begingroup$ Thank You. Precise and to the Point. $\endgroup$
    – Alexander.
    Commented Oct 18, 2014 at 8:32
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Yes, your method is correct.

Lets call the distance from $(2, 4)$ to $(0, 3)$, $r$. This is the maximux radius for the circle, because it's the largest size for the circle which hasn't $(0, 3)$ on the circle.

If you prove that the distance of $(2, 4)$ to $(3, 1)$ is larger than $r$, it will also proves that $(3, 1)$ is not on the circle.

In a nutshell:

Let's say that the points are $A = (2, 4), B = (3, 0), C = (3, 1)$

$d(A, B) > r$, because point $B$ isn't on the circle. If $d(A, C) > d(A, B)$ then: $d(A, C) > r$

So $(3, 1)$ is not on the circle.

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