Problem 4.2
We're told that y = 2x is the equation of the tangent line through point Q.
Plug this into the equation of the circle (that was given at the very top of the page). Solve for x.
x^2 + y^2 - 18x - 6y + 45 = 0
x^2 + (2x)^2 - 18x - 6(2x) + 45 = 0 ... every "y" replaced with "2x"
x^2 + 4x^2 - 18x - 12x + 45 = 0
5x^2 - 30x + 45 = 0
5(x^2 - 6x + 9) = 0
5(x-3)^2 = 0
(x-3)^2 = 0
x-3 = 0
x = 3
This then means y = 2x = 2*3 = 6
The coordinates of point Q are (x,y) = (3,6)
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Answer: (3,6)
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Problem 4.3
We first need to find the location of point S. That will involve solving this system of equations
[tex]\begin{cases}x-2y+12=0\\y = 2x\end{cases}[/tex]
which represent the equations of lines SR and QS in that exact order.
Like before, we'll apply substitution to solve for x.
x-2y+12 = 0
x-2(2x)+12 = 0
x-4x+12 = 0
-3x+12 = 0
12 = 3x
12/3 = x
4 = x
x = 4
Which leads to y = 2x = 2*4 = 8
Point S is located at (4,8)
Point R is given to be located at (6,9)
Apply the distance formula for these two points to find the length of SR (aka the distance from S to R).
[tex]S = (x_1,y_1) = (4,8)\\\\R = (x_2,y_2) = (6,9)\\\\d = \text{Length of SR, aka distance from S to R}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(4-6)^2 + (8-9)^2}\\\\d = \sqrt{(-2)^2 + (-1)^2}\\\\d = \sqrt{4 + 1}\\\\d = \sqrt{5}\\\\d \approx 2.236068\\\\[/tex]
The exact length is [tex]\sqrt{5}[/tex] and that approximates to roughly 2.236068 units.
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Answer:
Exact length = [tex]\sqrt{5}[/tex] units
Approximate length = 2.236068 units
