Question

If (8, −9) and (6, −3) are the endpoints of the diameter of a circle, what is the equation of the circle?

Answer 1

$(x-7)^{2} + (y+6)^{2} = 10$ is the equation of the required circle.

Step-by-step explanation:

Step 1 :

Let A = (8,-9) and B = (6,-3) be the endpoints of the diameter of the given circle.

We need to determine the circle's equation with this diameter

Step 2 :

The mid point of the diameter will be center of the required circle.

Let M = (x,y) be the mid point of the circle.

So  we have x = (8+6)÷ 2 , y = ((-9) + (-3)) ÷ 2

M = ( 7, -6) is the midpoint of the diameter and the center of the circle.

Step 3 :

The radius of the circle is  distance between  midpoint and any one of the end point of the diameter that is  distance between  point A(8,-9) and  M ( 7,-6)

Distance between the 2 points $(x_{1} ,y_{1}) and (x_{2} ,y_{2})$ is given by

$\sqrt{(x_{2}-x_{1})^{2}+ (y_{2}-y_{1})^{2} }$

So distance between  A(8,-9) and  M ( 7,-6)  =  $\sqrt{(7-8)^{2}+ (-6-(-9))^{2} }$

                                                                        =  $\sqrt{(1)^{2}+ (3)^{2} } = \sqrt{10 }$

Hence the radius r is $\sqrt{10}$

Step 4 :

The equation of circle with center (a,b) = (7,-6) and radius r = $\sqrt{10}$ is given by

$(x-a)^{2} + (y-b)^{2} = r^{2}$

=> $(x-7)^{2} + (y-(-6))^{2} = (\sqrt{10} )^{2}$

=> $(x-7)^{2} + (y+6)^{2} = 10$ is equation of the required circle.

Step 5 :

Answer :

$(x-7)^{2} + (y+6)^{2} = 10$  is  equation of the required circle.