# Prove that point M is on circle c

It's hard to create question names that make sense. Anyhow, the following is another question from my math assignment.

Line-segment AB has a fixed length of 10 units. point A moves on the positive x-axis and point B moves on the positive y-axis.

Point M is in the middle of line AB.

All right, I have been able to figure out the first question of the assignment: that M always stays on a circle from the Origin with a radius of 5.

However, the second question asks me to prove my answer from the first question.

How do I prove that M will always stay on this circle?

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This won't help you on your quest for an answer, but when cutting glass you can use a setup like this. You need two perpendicular rails on which $A$ and $B$ can move freely (past the origin too), and you put a cutter at $M$. If you put $M$ somewhere on $AB$ other than the midpoint (doesn't even have to be between $A$ and $B$), then $M$ will follow an ellipse. –  Arthur Nov 17 '12 at 15:43

To prove that $M$ always stays on the circle with center $O$, and radius 5, you have to prove that $OM = 5$, no matter where $A, B$ are.

There is a theorem on right triangle, saying that: "The median on the hypotenuse of a right triangle equals one-half the hypotenuse", and remember that $AB = 10$, can you prove $OM = 5$?

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Let $M(h,k), A(a,0)$ and $B(0,b)$

So, using Pythagoras Theorem, $a^2+b^2=10^2$

As $M$ is the midpoint, $$h=\frac{a+0}2,k=\frac{0+b}2\implies a=2h,b=2k$$

So, $(2h)^2+(2k)^2=10^2\implies h^2+k^2=5^2$

So, the locus of $M(h,k)$ is $x^2+y^2=25$ which is evidently a circle centred at the origin with radius $5$ unit.

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You should compute the analytical representation of the locus defined by your definition of M.

A = (x,0)
B = (0,sqrt(10^2-x^2))

M = (x/2, sqrt(10^2-x^2)/2)

Circle centered in 0 with radius 5: x^2+y^2 = 5^2


To prove that M is always on the circle you should substitute its x and y coordinates in the equation of the circle and observe that you obtain an identity.

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