Answer to Question #234697 in Analytic Geometry for Johnnie

Question

Two points A and B have coordinates (3,6) and (-3,0) respectively.

Obtain the equation of the circle for which the line segment AB is the diameter

Answer 1

The general equation for a circle is $( x - h )^2 + ( y - k )^2 = r^2,$ where ($h, k )$ is the center and $r$ is the radius.

The line segment $AB$ is the diameter. Then the center of the circle $C(h, k)$ is the middle of the line segment $AB.$

$$h=\dfrac{x_A+x_B}{2}=\dfrac{3-3}{2}=0$$

$$k=\dfrac{y_A+y_B}{2}=\dfrac{6+0}{2}=3$$

Center $C(0,3).$

The radius of the circle is half of the diameter

$$r=\dfrac{\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}}{2}$$

$$=\dfrac{\sqrt{(-3-3)^2+(0-6)^2}}{2}=3\sqrt{2}$$

Substitute

$$( x - 0 )^2 + ( y - 3 )^2 = (3\sqrt{2})^2$$

The general equation of the circle  for which the line segment $AB$ is the diameter is 

$$x^2 + ( y - 3 )^2 = 18$$