Question

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

Answer 1

Centre of the circle lies at : `(x,y)=(3,5)`

Explanation:

Suppose if `\vec{r_A}=(x_A, y_A)` and `\vec{r_B}=(x_B, y_B)` 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 : `\vec{r} = \vec{r_A}+\gamma(\vec{r_B}-\vec{r_B})`
X Component: `x = x_A + \gamma (x_B-x_A)`;
Y Component: `y = y_A + \gamma (y_B-y_A)`;

If `d` is the distance of point C from point A and `D` the distance between points A and B, then `\gamma = 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=D/2`.

So `\gamma=d/D =1/2`

So the coordinates of the centre `(x,y)` are :
`x = x_A+1/2(x_B-x_A)=7+1/2(-1-7)=3`;
`y = y_A+1/2(y_B-y_A)=2+1/2(8-2)=5`

Therefore `(x,y)=(3,5)` 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_{AC}=\sqrt{(x-x_A)^2=(y-y_A)^2}`
`\qquad=\sqrt{(3-7)^2+(5-2)^2}=5`
B-C Distance:
`d_{BC}=\sqrt{(x-x_B)^2=(y-y_B)^2}`
`\qquad=\sqrt{(3-(-1))^2+(5-8)^2}=5`