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

Answer:

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#