Question

a circle has a diameter with endpoints at 15 + 25i and -25 - 17i. which point is also on the circle? -15 + 21i 0 + 0i 15 + 17i 16 + 24i

Explanation:

Step1: Find the center of the circle

The center of the circle (a + bi) with diameter endpoints \(z_1=x_1 + y_1i\) and \(z_2=x_2 + y_2i\) is \(\frac{z_1 + z_2}{2}\). Here \(z_1=15 + 25i\), \(z_2=-25-17i\). Then \(a+bi=\frac{(15 + 25i)+(-25-17i)}{2}=\frac{(15-25)+(25 - 17)i}{2}=\frac{-10 + 8i}{2}=-5+4i\).

Step2: Find the radius of the circle

The radius \(r\) is the distance from the center to an endpoint. Using the distance formula for complex - numbers \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Let the center \(z_0=-5 + 4i\) and \(z_1=15 + 25i\). Then \(r=\sqrt{(15+5)^2+(25 - 4)^2}=\sqrt{20^2+21^2}=\sqrt{400 + 441}=\sqrt{841}=29\).

Step3: Check each point

For a point \(z=x+yi\) on the circle, the distance from the center \(z_0=-5 + 4i\) to \(z\) should be equal to \(r\).
For \(z=-15 + 21i\):
The distance \(d=\sqrt{(-15 + 5)^2+(21 - 4)^2}=\sqrt{(-10)^2+17^2}=\sqrt{100+289}=\sqrt{389}
eq29\).
For \(z = 0+0i\):
The distance \(d=\sqrt{(0 + 5)^2+(0 - 4)^2}=\sqrt{25 + 16}=\sqrt{41}
eq29\).
For \(z=15 + 17i\):
The distance \(d=\sqrt{(15 + 5)^2+(17 - 4)^2}=\sqrt{20^2+13^2}=\sqrt{400+169}=\sqrt{569}
eq29\).
For \(z = 16+24i\):
The distance \(d=\sqrt{(16 + 5)^2+(24 - 4)^2}=\sqrt{21^2+20^2}=\sqrt{441+400}=\sqrt{841}=29\). So the point \(16 + 24i\) is on the circle.

Answer:

A. -15 + 21i