Finding the variable of a coordinate point on a circle

Question

This might be a very simple question but I am having trouble figuring it out, so if anyone can explain: A circle is marked with three points A(-3,2),B(9,10) and C(x,4). All three points lie on the circle. Given that AC is a diameter of the circle how would you find the value of the variable x?

Answer 1

If $AC$ is a diameter then $\widehat{ABC}=\frac{\pi}{2}$, hence $BC$ is orthogonal to $AB$ and we can find the abscissa of $C$ by intersecting the $y=4$ line with the perpendicular to $AB$ through $B$, whose equation is $y=-\frac{3}{2}x+\frac{47}{2}$. Setting $4=-\frac{3}{2}x+\frac{47}{2}$ gives $x=13$.

Answer 2

Since $\overline{AC}$ is a diameter of the circle, its midpoint $M(\frac{x-3}{2},3)$ is the center. From the definition of the circle, it follows that $|BM|^2=|CM|^2$ i.e.

$$\hspace{12mm}\left(9-\frac{x-3}{2}\right)^2+(10-3)^2=\left(x-\frac{x-3}{2}\right)^2+(4 -3)^2$$ \begin{align} \implies \left(9-\frac{x-3}{2}\right)^2-\left(x-\frac{x-3}{2}\right)^2 &=12(9-x)\\&=(4 -3)^2-(10-3)^2\\&=-6(8)\implies \fbox{$x=13$} \end{align}

Answer 3

If AC is a diameter of the circle, then $\angle B = \dfrac {π}{2}$.

Therefore, $m_{AB} . m_{BC} = –1$. That is, $\dfrac {10 - 2}{9 - -3} . \dfrac {10 - 4}{9 - x} = -1$.

$x = ... = 13$.