A circle has a diameter with endpoints at –2 + i and –6 –11i. What is the center of the circle?
1 answer:
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<h2>Answer:</h2>
4π
<h2>Step-by-step explanation:</h2>As shown in the diagram, triangle XYZ is a right triangle. Therefore, its area (A) is given by:
A = x b x h -------------(i)
Where;
A = 18
b = XZ = base of the triangle
h = YZ = height of the triangle = 6
<em>Substitute these values into equation(i) and solve as follows:</em>
18 =
x b x 6
18 = 3b
<em>Divide through by 3</em>
6 = b
Therefore, b = XZ = 6
<em>Now, assume that the circle is centered at O;</em>
Triangle XOZ is isosceles, therefore the following are true;
(i) |OZ| = |OX|
(ii) XZO = ZXO = 30°
(iii) XOZ + XZO + ZXO = 180° [sum of angles in a triangle]
=> XOZ + 30° + 30° = 180°
=> XOZ + 60° = 180°
=> XOZ = 180° - 60°
=> XOZ = 120°
Therefore we can calculate the radius |OZ| of the circle using sine rule as follows;
|OZ| = 6
The radius of the circle is therefore 6.
<em>Now, let's calculate the length of the arc XZ</em>
The length(L) of an arc is given by;
L = θ / 360 x 2 π r ------------------(ii)
Where;
θ = angle subtended by the arc at the center.
r = radius of the circle.
In our case,
θ = ZOX = 120°
r = |OZ| = 6
Substitute these values into equation (ii) as follows;
L = 120/360 x 2π x 6
L = 4π
Therefore the length of the arc XZ is 4π
Hi,
Work:
Equation;
Root = (0, 0)
Domain: x = R
Vertical Intercept = (0, 0)
Hope this helps.
r3t40
Work:
Equation;
Root = (0, 0)
Domain: x = R
Vertical Intercept = (0, 0)
Hope this helps.
r3t40