Answer to Question #234697 in Analytic Geometry for Johnnie

Question #234697

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


1
Expert's answer
2021-09-09T00:40:39-0400

The general equation for a circle is (xh)2+(yk)2=r2,( x - h )^2 + ( y - k )^2 = r^2, where (h,k)h, k ) is the center and rr is the radius.

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


h=xA+xB2=332=0h=\dfrac{x_A+x_B}{2}=\dfrac{3-3}{2}=0

k=yA+yB2=6+02=3k=\dfrac{y_A+y_B}{2}=\dfrac{6+0}{2}=3

Center C(0,3).C(0,3).


The radius of the circle is half of the diameter


r=(xBxA)2+(yByA)22r=\dfrac{\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}}{2}

=(33)2+(06)22=32=\dfrac{\sqrt{(-3-3)^2+(0-6)^2}}{2}=3\sqrt{2}

Substitute


(x0)2+(y3)2=(32)2( x - 0 )^2 + ( y - 3 )^2 = (3\sqrt{2})^2



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


x2+(y3)2=18x^2 + ( y - 3 )^2 = 18


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