# If the diameter of a circle has endpoints A(7, 2) and B(-1, 8), where is the center of the circle?

##### 1 Answer
Feb 28, 2016

Centre of the circle lies at : $\left(x , y\right) = \left(3 , 5\right)$

#### Explanation:

Suppose if $\setminus \vec{{r}_{A}} = \left({x}_{A} , {y}_{A}\right)$ and $\setminus \vec{{r}_{B}} = \left({x}_{B} , {y}_{B}\right)$ are the position vectors of any two points A and B in a plane space, then the position vector of any point C along the line connecting the two points can be written as a parametric equation : $\setminus \vec{r} = \setminus \vec{{r}_{A}} + \setminus \gamma \left(\setminus \vec{{r}_{B}} - \setminus \vec{{r}_{B}}\right)$
X Component: $x = {x}_{A} + \setminus \gamma \left({x}_{B} - {x}_{A}\right)$;
Y Component: $y = {y}_{A} + \setminus \gamma \left({y}_{B} - {y}_{A}\right)$;

If $d$ is the distance of point C from point A and $D$ the distance between points A and B, then $\setminus \gamma = \frac{d}{D}$

Solution: We are given -
\vec{r_A}=(x_A, y_A)=(7,2); \qquad \vec{r_B}=(x_B, y_B)=(-1, 8).

If A and B are the end points along the diameter then we know that the centre of the circle is the midpoint between the two. i.e $d = \frac{D}{2}$.

So $\setminus \gamma = \frac{d}{D} = \frac{1}{2}$

So the coordinates of the centre $\left(x , y\right)$ are :
$x = {x}_{A} + \frac{1}{2} \left({x}_{B} - {x}_{A}\right) = 7 + \frac{1}{2} \left(- 1 - 7\right) = 3$;
$y = {y}_{A} + \frac{1}{2} \left({y}_{B} - {y}_{A}\right) = 2 + \frac{1}{2} \left(8 - 2\right) = 5$

Therefore $\left(x , y\right) = \left(3 , 5\right)$ is the centre.

Note: You can verify the correctness of the solution by finding the distance between the centre and the two points and verify that they are at the same distance.
A-C Distance:
${d}_{A C} = \setminus \sqrt{{\left(x - {x}_{A}\right)}^{2} = {\left(y - {y}_{A}\right)}^{2}}$
$\setminus q \quad = \setminus \sqrt{{\left(3 - 7\right)}^{2} + {\left(5 - 2\right)}^{2}} = 5$
B-C Distance:
${d}_{B C} = \setminus \sqrt{{\left(x - {x}_{B}\right)}^{2} = {\left(y - {y}_{B}\right)}^{2}}$
$\setminus q \quad = \setminus \sqrt{{\left(3 - \left(- 1\right)\right)}^{2} + {\left(5 - 8\right)}^{2}} = 5$