I have three points$A(-10,-12)$,$B(6,18)$,$C(-2,-14)$. I have to prove they exist on the same semicircle. How?
If I try to draw the three points to investigate I see that I can draw a right triangle. I know you can inscribe a right triangle between three points of a semicircle, the diameter being the hypotenuse of this triangle.
So I thought I could do Pythagoras/distance formula between the points to find $AB$, $BC$, and $AC$. Then verify with Pythagoras whether $\sqrt{AC^2+BC^2}=BA$ , but for some reason this does not add up to the expected result (why not?)
So I looked up the solution and it reads:\begin{align*} m_{AB} &= \frac{y_B-y_A}{x_B-x_A} = \phantom{-}\frac{18+12}{6+10} = \frac{30}{16} = \frac{15}{8} \\ m_{BC} &= \frac{y_C-y_B}{x_C-x_B} = \frac{-14-18}{-2-6} = \frac{-32}{-8} = 4 \\ m_{CA} &= \frac{y_A-y_C}{x_A-x_C} = \frac{-12+14}{-10+2} = \frac{2}{-8} = -\frac{1}{4} \end{align*}
Since$$ 4 \cdot \frac{-1}{4} = -1, $$$BC$ is perpendicular to $CA$ (negative reciprocal) and hence $\triangle ABC$ is a right angled triangle.
Since the angle in a semicircle is a right angle it follows that $AB$ is a diameter and the points form a semicircle.
So I understand that
$$m_{AB} = \frac{y_B- y_A}{x_B-x_A}$$
Is the gradient formula$$ \frac{\text{rise}}{\text{run}} $$
So basically this (more efficient) solution is ignoring lengths/distances and leapfrogging lines and is going straight to prove there is a right angle (clever!)
But I wonder... why would my solution still not work?
$\endgroup$ 42 Answers
$\begingroup$We have $\vec{AC}=(8,-2)$, $\vec{BC}=(-8,-32)$ and $\vec{BA}=(-16,-30)$
so we have
$$|AC|^2+|BC|^2=68+1088=1156=|BA|^2$$
Thus $\frac{1}{2}|BA|=17$ is the radius of the circle, centred at $(-2,3)$ and thus its equation is given by
$$(x+2)^2+(y-3)^2=17^2$$
For three points of a plane passes only one circumference, and this has equation:
$x^{2}+y^{2}+4x-6y-276=0$.
In fact, the coordinates of the three points satisfy the equation, that is, they make it an identity.
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