How do I find the 3rd point on a circle to make an isosceles triangle?

Question

I'm trying to find the point C in the following diagram:

isosceles missing point

I know the coordinates for points A and B and also the centre of the circle. I also know the height of the isosceles triangle (h) which is also the radius of the circle.

How do I go about finding the coordinates for point C? I understand that there will be two answers, each on opposite sides of the circle.

Answer 1

I’ll refer to the coordinates of $A$ as $A(x_A,y_A)$ and I’ll refer to the coordinates of all other points in the same way ($C(x_C,y_C)$,$O(x_O,y_O)$ , etc.)

Since $\overline{AB}$ touches $O$, or the center of the circle, then $\overline{AB}$ must be a diameter of $\odot O$

Since $h$ is the height of $\vartriangle{ABC}$ and is also the length of $\overline{OC}$ then $\overline{AB}\bot \overline{OC}$

Because they are all radii $\overline{AO}\cong \overline{CO} \cong \overline{OB}$.

That means that $C$ is just a $90^{\circ}$ rotation in either direction of $A$ around $O$

To compute that point, you have to find the change in x and change in y between $A$ and $O$. These are as follows: $$\Delta{x} = x_O-x_A$$ $$\Delta{y} = y_O-y_A$$

The two possible points for $C(x_C,y_C)$ are then:

$$C(x_O+\Delta{y},y_O-\Delta{x})$$ $$C(x_O-\Delta{y},y_O+\Delta{x})$$

Answer 2

If $A= (x_a,y_a)$ and $B=(x_b, y_b)$ then the slope of the line $AB$ is $m = \frac{x_b-x_a}{y_b-y_a}$. So the slope of a line perpendicular to $AB$ has a slope $-\frac 1m = \frac {y_a-y_b}{x_b - x_a}$.

$AB$ is clearly a diameter of the circle (otherwise the height of the triangle cant be equal to the radius.

So the center point is $(\frac {x_a + x_b}2, \frac{y_a+y_b}2)=(x_o,y_o)=O$. And the perpencular bisector of $AB$ will by $y = -\frac 1m(x- x_o)+ y_o$.

We need 1: $C=(x_c,y_c)$ so that $C$ is on the perpendicular bisector so $AC=BC$. That means:

$y_c =-\frac 1m(x_c-x_o) + y_o$

And the distance of $CO$ which is $\sqrt{(y_c - y_o)^2 - (x_c- x_a)^2}=$

must be equal to $h$.

But $h = OA = OB = \frac {AB}2 = \frac {\sqrt{(y_b-y_a)^2-(x_b-x_a)^2}}2$

So we must have $\sqrt{(y_c - y_o)^2 - (x_c- x_a)^2}=\sqrt{(y_a - y_o)^2 +(x_a-x_o)^2}$

Sustituting $y_c =-\frac 1m(x_c-x_o) + y_o$ we have

$((-\frac 1m(x_c-x_o) + y_o - y_o)^2 - (x_c- x_a)^2=(y_a - y_o)^2 +(x_a-x_o)^2$

or $|\frac 1m(x_c-x_o)|=|y_a-y_o|$ or $x_c =\pm m(y_a-y_o)+x_o$

And solving for $y_c$ we have =

$y_c=-\frac 1m(x_c-x_o) + y_o= -\frac 1m(\pm m(y_a-y_o)+x_o-x_o))=\pm (y_a-y_o)$.