Question

Find the coordinates of a point $A$, where $AB$ is the diameter of a circle whose center is $(2,-3)$ and $B$is $(1,4)$.

Answer 1

Solve for the coordinates of the point $A$

Given,

$B(x_1,y_1)=(1,4)$

Let $C$ be the center of the circle, then

$C(x,y)=(2,-3)$

Let $\left(x_2,y_2\right)$ be the coordinates of the point $A$

Also given, $AB$ is the diameter of the circle.

If $AB$ is the diameter of the circle and $C$ is the center of the circle, then by the definition of a diameter, we can say that $C$ is the midpoint of the line $AB$.

Thus, we can use the midpoint formula to solve for the coordinates of the point.

For two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$, the midpoint $P$ is given by the formula,

$\mathit{P}\mathbf{(}\mathit{x}\mathbf{,}\mathbf{}\mathit{y}\mathbf{)}\mathbf{=}\mathbf{(}\frac{{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{x}}_{\mathbf{1}}}{\mathbf{2}}\mathbf{,}\frac{{\mathbf{y}}_{\mathbf{2}}\mathbf{+}{\mathbf{y}}_{\mathbf{1}}}{\mathbf{2}}\mathbf{)}$

We know the midpoint and one endpoint of the line, thus, in this case,

$$\begin{gathered} \left(x,y\right)=\left(2,-3\right) \\ \left(x_1,y_1\right)=\left(1,4\right) \end{gathered}$$

So, by using the formula

$\begin{array}{rcl}(2,-3)& =& \left(\frac{1+x_2}{2},\frac{4+y_2}{2}\right)\end{array}$

Comparing the coordinates,

$\begin{array}{rcl}& & \begin{array}{cc}& \frac{1+x_2}{2}=2\\ \Rightarrow & 1+x_2=4\\ \Rightarrow & x_2=3\end{array}\end{array}$

$\begin{array}{cc}& \frac{4+y_2}{2}=-3\\ \Rightarrow & 4+y_2=-6\\ \Rightarrow & y_2=-10\end{array}$

Hence, the coordinates of the point $A$ is $(3,-10)$.