If the diameter of a circle is segment AB where Point A is located at (-3, 6), and Point B located at (-8, -1), what is the diameter of the circle?
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Answer:
√74 units.
Explanation:
Given that segment AB is the diameter of the circle and the points A and B are located at:
[tex]\begin{gathered} (x_1,y_1)=A(-3,6) \\ \left(x_2,y_2\right)=B\left(-8,-1\right) \end{gathered}[/tex]The diameter of the circle is the length of segment AB.
To find the length of the segment, we use the distance formula given below:
[tex]$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} $[/tex]Substitute the given values:
[tex]\begin{gathered} AB=\sqrt{(-8-\left(-3\right))^2+(-1-6)^2} \\ =\sqrt{(-8+3\rparen^2+(7)^2} \\ =\sqrt{\lparen-5)^2+7^2} \\ =\sqrt{25+49} \\ \implies AB=\sqrt{74} \end{gathered}[/tex]The diameter of the circle is √74 units.
The last option is correct.