Question

a circle has a diameter with endpoints at -3 + 18i and 7 - 6i. which point is also on the circle?
-7 + 7i
5 + 12i
12 - i
14 + 11i

Explanation:

Step1: Find the center of the circle

The center of the circle is the mid - point of the diameter. If the endpoints of the diameter are \(z_1=-3 + 18i\) and \(z_2=7-6i\), the mid - point formula for complex numbers \(z_m=\frac{z_1 + z_2}{2}\). So \(z_m=\frac{(-3 + 18i)+(7-6i)}{2}=\frac{(-3 + 7)+(18i-6i)}{2}=\frac{4 + 12i}{2}=2+6i\).

Step2: Find the radius of the circle

The radius \(r\) is the distance between the center \(z_m = 2+6i\) and one of the endpoints, say \(z_1=-3 + 18i\). The distance formula for complex numbers \(d=\vert z_2 - z_1\vert=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here \(r=\vert(-3 + 18i)-(2+6i)\vert=\vert(-3-2)+(18i - 6i)\vert=\vert-5 + 12i\vert=\sqrt{(-5)^2+12^2}=\sqrt{25 + 144}=\sqrt{169}=13\).

Step3: Check each point

For a point \(z=x+yi\) to be on the circle, the distance between \(z\) and the center \(z_m = 2+6i\) should be equal to the radius.

• For \(z=-7 + 7i\): \(\vert(-7 + 7i)-(2+6i)\vert=\vert-9 + i\vert=\sqrt{(-9)^2+1^2}=\sqrt{81 + 1}=\sqrt{82}

eq13\).

• For \(z = 5+12i\): \(\vert(5 + 12i)-(2+6i)\vert=\vert(5 - 2)+(12i-6i)\vert=\vert3 + 6i\vert=\sqrt{3^2+6^2}=\sqrt{9 + 36}=\sqrt{45}

eq13\).

• For \(z=12 - i\): \(\vert(12 - i)-(2+6i)\vert=\vert(12 - 2)+(-i-6i)\vert=\vert10-7i\vert=\sqrt{10^2+(-7)^2}=\sqrt{100 + 49}=\sqrt{149}

eq13\).

• For \(z = 14+11i\): \(\vert(14 + 11i)-(2+6i)\vert=\vert(14 - 2)+(11i-6i)\vert=\vert12 + 5i\vert=\sqrt{12^2+5^2}=\sqrt{144 + 25}=\sqrt{169}=13\). So the point \(14 + 11i\) is on the circle.

Answer:

B. \(5 + 12i\)