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 : (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