Let A and B be two different points. Show that the points P are such that the angle APB is 90 degrees and creates a circle. Decide the the radius and mid point of the circle.
I have problems proving that the angle have to be 90 degrees, isnt it only 90 degrees if the base of the triangle in the circle is the diagonal of the circle?
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Without loss of generality, let A be (0,0) and let B = (a,b). Then |AB| = $\sqrt{a^2+b^2}$.
We are asking for the set of points P(x,y) such that $\angle$APB is a right angle.
Suppose P is such a point. Then |PA| = $\sqrt{x^2+y^2}$ and |PB| = $\sqrt{(x-a)^2+(y-b)^2}$.
So $\triangle$APB is a right triangle. Then by the Pythagorean theorem:
$|PA|^2+|PB|^2 = |AB|^2$
$(x^2+y^2)+(x-a)^2+(y-b)^2 = (a^2+b^2)$
If we expand the left-hand side and complete the square, we get
$\left(x-\frac{a}{2}\right)^2+\left(y-\frac{b}{2}\right)^2 = \left(\frac{a^2+b^2}{4}\right)$.
This equation is the equation of a circle with the midpoint of |AB| as the center with radius $\frac{|AB|}{2}$.
If $C$ is midpoint of $AB$ and $\angle APB$ is a right angle, draw a line thru $C$ parallel to $AP$. That line will divide $PB$ in half at point $D$ (midsection theorem). It will also be perpendicular to $PB$ (because $AP$ is perpendicular to $PB$). Similarly, draw a line thru $C$ parallel to $BP$ and see that that line is perpendicular bisector to $AP$. Thus, $C$ is a center of the circumcenter.
Conversely, if you have that $CP=CA=CB$, use sum of angles to show that $\alpha+\beta=90°$