Solutions to Problem Set 26

Math 101-06

11-13-2017

[Circles]

1. Find the distance between the points $(3, -6)$ and $(4, 1)$ .

$$\sqrt{(3 - 4)^2 + (-6 - 1)^2} = \sqrt{50} = 5 \sqrt{2}.\quad\halmos$$


2. Write down the standard equation of the circle with radius 5 and center $(9, -7)$ .

$$(x - 9)^2 + (y + 7)^2 = 25.\quad\halmos$$


3. Write down the standard equation of the circle with radius $\sqrt{3}$ and center $(2,
   4)$ .

$$(x - 2)^2 + (y - 4)^2 = 3.\quad\halmos$$


4. Find the standard equation, the center, and the radius of the circle

$$x^2 + 6 x + y^2 - 4 y = 3.$$

$$\eqalign{ x^2 + 6 x + + 9 + y^2 - 4 y + 4 & = 3 + 9 + 4 \cr (x + 3)^2 + (y - 2)^2 & = 16 \cr}$$

Center: $(-3, 2)$ . Radius: 4.


5. Find the standard equation, the center, and the radius of the circle

$$x^2 - 2 x + y^2 - 8 y - 32 = 0.$$

$$\eqalign{ x^2 - 2 x + 1 + y^2 - 8 y + 16 - 32 & = 0 + 1 + 16 \cr (x - 1)^2 + (y - 4)^2 - 32 & = 17 \cr (x - 1)^2 + (y - 4)^2 - 32 + 32 & = 17 + 32 \cr (x - 1)^2 + (y - 4)^2 & = 49 \cr}$$

Center: $(1, 4)$ . Radius: 7.


6. Use the Quadratic Formula to solve

$$x^2 + 8 x - 10 = 0.$$

$$x = \dfrac{-8 \pm \sqrt{8^2 - 4(1)(-10)}}{2 \cdot 1} = \dfrac{-8 \pm \sqrt{64 + 40}}{2} = \dfrac{-8 \pm \sqrt{104}}{2} = \dfrac{-8 \pm 2 \sqrt{26}}{2} = -4 \pm \sqrt{26}.\quad\halmos$$


7. Use the Quadratic Formula to solve

$$x^2 + 2 x + 4 = 0.$$

$$x = \dfrac{-2 \pm \sqrt{2^2 - 4(1)(4)}}{2 \cdot 1} = \dfrac{-2 \pm \sqrt{4 - 16}}{2} = \dfrac{-2 \pm \sqrt{-12}}{2} = \dfrac{-2 \pm 2 i \sqrt{3}}{2} = -1 \pm i \sqrt{3}.\quad\halmos$$


Courage consists of the power of self-recovery. - Ralph Waldo Emerson


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