Question #156836
Find the equation of a circle when the 2 end points of a diameter is given by A(-2, 1) , B(-1, 2)
1
Expert's answer
2021-02-02T04:30:01-0500

The center of the circle has the coordinates: O(32,32)O(-\frac32,\frac32). It is the midpoint of the segment A(-2,1), B(-1,2). The radius of the circle is: r=(32+2)2+(321)2=12=22r=\sqrt{(-\frac32+2)^2+(\frac32-1)^2}=\sqrt{\frac12}=\frac{\sqrt{2}}{2} . The equation is (x+32)2+(y32)2=12(x+\frac32)^2+(y-\frac32)^2=\frac12 .

We can also consider the problem with complex numbers. I.e., we may present points in the following form : A:-2+i; B:-1+2i and O: 32+32i-\frac32+\frac32i. The radius is r=22r=\frac{\sqrt{2}}{2} (the distance between O and A). The equation of the circle is: zz02=12|z-z_0|^2=\frac12 , where zz presents a complex variable. I.e., z=x+iyz=x+iy and z0=32+32iz_0=-\frac32+\frac32i


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