2. Add: Simplify and state the domain.
2 answers:
Answer:
see explanation
Step-by-step explanation:
+
← factor the denominators
= +
= +
← simplify numerator
=
=
The denominator cannot be zero as this would make the rational function undefined
Equating the denominator to zero and solving gives the values of x that cannot be in the domain
5(x + 2)(x - 4) = 0
Equate each factor to zero and solve for x
x + 2 = 0 ⇒ x = - 2
x - 4 = 0 ⇒ x = 4
x = - 2, x = 4 ← excluded values
domain : x ∈ R ( x ≠ - 2, x ≠ 4 )
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<h3>Explanation:</h3>
1. "Create your own circle on a complex plane."
The equation of a circle in the complex plane can be written a number of ways. For center c (a complex number) and radius r (a positive real number), one formula is ...
|z-c| = r
If we let c = 2+i and r = 5, the equation becomes ...
|z -(2+i)| = 5
For z = x + yi and |z| = √(x² +y²), this equation is equivalent to the Cartesian coordinate equation ...
(x -2)² +(y -1)² = 5²
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2. "Choose two end points of a diameter to prove the diameter and radius of the circle."
We don't know what "prove the diameter and radius" means. We can show that the chosen end points z₁ and z₂ are 10 units apart, and their midpoint is the center of the circle c.
For the end points of a diameter, we choose ...
- z₁ = 5 +5i
- z₂ = -1 -3i
The distance between these is ...
|z₂ -z₁| = |(-1-5) +(-3-5)i| = |-6 -8i|
= √((-6)² +(-8)²) = √100
|z₂ -z₁| = 10 . . . . . . the diameter of a circle of radius 5
The midpoint of these two point should be the center of the circle.
(z₁ +z₂)/2 = ((5 -1) +(5 -3)i)/2 = (4 +2i)/2 = 2 +i
(z₁ +z₂)/2 = c . . . . . the center of the circle is the midpoint of the diameter
__₁₂₃₄
3. "Show how to determine the center of the circle."
As with any circle, the center is the <em>midpoint of any diameter</em> (demonstrated in question 2). It is also the point of intersection of the perpendicular bisectors of any chords, and it is equidistant from any points on the circle.
Any of these relations can be used to find the circle center, depending on the information you start with.
As an example. we can choose another point we know to be on the circle:
z₄ = 6-2i
Using this point and the z₁ and z₂ above, we can write three equations in the "unknown" circle center (a +bi):
- |z₁ - (a+bi)| = r
- |z₂ - (a+bi)| = r
- |z₄ - (a+bi)| = r
Using the formula for the square of the magnitude of a complex number, this becomes ...
(5-a)² +(5-b)² = r² = 25 -10a +a² +25 -10b +b²
(-1-a)² +(-3-b)² = r² = 1 +2a +a² +9 +6b +b²
(6-a)² +(-2-b)² = r² = 36 -12a +a² +4 +4b +b²
Subtracting the first two equations from the third gives two linear equations in a and b:
11 -2a -21 +14b = 0
35 -14a -5 -2b = 0
Rearranging these to standard form, we get
a -7b = -5
7a +b = 15
Solving these by your favorite method gives ...
a +bi = 2 +i = c . . . . the center of the circle
__
4. "Choose two points, one on the circle and the other not on the circle. Show, mathematically, how to determine whether or not the point is on the circle."
The points we choose are ...
- z₃ = 3 -2i
- z₄ = 6 -2i
We can show whether or not these are on the circle by seeing if they satisfy the equation of the circle.
|z -c| = 5
For z₃: |(3 -2i) -(2 +i)| = √((3-2)² +(-2-i)²) = √(1+9) = √10 ≠ 5 . . . NOT on circle
For z₄: |(6 -2i) -(2 +i)| = √((6 -2)² +(2 -i)²) = √(16 +9) = √25 = 5 . . . IS on circle