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arsen [322]
3 years ago
10

Which equation represents the circle described?

Mathematics
2 answers:
faust18 [17]3 years ago
3 0
ANSWER

{(x - 4)}^{2} +  {(y - 3)}^{2}  =   {2}^{2}

EXPLANATION

The equation of the circle with radius r and centre (a,b) is given by


(x - a)^{2}  +  {(x - b)}^{2}  =  {r}^{2}

The radius is

r = 2

We need to determine the center of the circle from the given equation of another circle, which is,


{x}^{2}  +  {y}^{2}  - 8x - 6y + 24 = 0


We complete the square to obtain,


{x}^{2}  - 8x+  {y}^{2}  - 6y + 24 = 0

{x}^{2}  - 8x+  {y}^{2}  - 6y  =  - 24
{x}^{2}  - 8x+  {( - 4)}^{2} +   {y}^{2}  - 6y  +  {( - 3)}^{2}  =  - 24 +  {( - 3)}^{2} +  {( - 4)}^{2}


{(x - 4)}^{2} +  {(y - 3)}^{2}  =  - 24 +  9+  16




{(x - 4)}^{2} +  {(y - 3)}^{2}  =  1


The centre of this circle is (4,3)


Hence the center of the circle whose equation we want to find is also (4,3).


With this center and radius 2, the required equation is,


{(x - 4)}^{2} +  {(y - 3)}^{2}  =   {2}^{2}


Therefore the correct answer is C.

bekas [8.4K]3 years ago
3 0

Answer:  the answer is C :)

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<h3>How to determine the dimensions?</h3>

The given parameters are:

Base = 5(x+3)

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So, we have:

2 *(5(x + 3) + 2(x + 9)) = 108

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Write the equation for a parabola with a focus at (-8, -1) and a directrix at y = − 4.
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x^2 + 16x + 64 + 1 - 16 = 8y - 2y

6y = x^2 + 16x + 49

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dezoksy [38]

We have to calculate the fourth roots of this complex number:

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We start by writing this number in exponential form:

\begin{gathered} r=\sqrt[]{9^2+(9\sqrt[]{3})^2} \\ r=\sqrt[]{81+81\cdot3} \\ r=\sqrt[]{81+243} \\ r=\sqrt[]{324} \\ r=18 \end{gathered}\theta=\arctan (\frac{9\sqrt[]{3}}{9})=\arctan (\sqrt[]{3})=\frac{\pi}{3}

Then, the exponential form is:

z=18e^{\frac{\pi}{3}i}

The formula for the roots of a complex number can be written (in polar form) as:

z^{\frac{1}{n}}=r^{\frac{1}{n}}\cdot\lbrack\cos (\frac{\theta+2\pi k}{n})+i\cdot\sin (\frac{\theta+2\pi k}{n})\rbrack\text{ for }k=0,1,\ldots,n-1

Then, for a fourth root, we will have n = 4 and k = 0, 1, 2 and 3.

To simplify the calculations, we start by calculating the fourth root of r:

r^{\frac{1}{4}}=18^{\frac{1}{4}}=\sqrt[4]{18}

<em>NOTE: It can not be simplified anymore, so we will leave it like this.</em>

Then, we calculate the arguments of the trigonometric functions:

\frac{\theta+2\pi k}{n}=\frac{\frac{\pi}{2}+2\pi k}{4}=\frac{\pi}{8}+\frac{\pi}{2}k=\pi(\frac{1}{8}+\frac{k}{2})

We can now calculate for each value of k:

\begin{gathered} k=0\colon \\ z_0=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{0}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{0}{2}))) \\ z_0=\sqrt[4]{18}\cdot(\cos (\frac{\pi}{8})+i\cdot\sin (\frac{\pi}{8}) \\ z_0=\sqrt[4]{18}\cdot e^{i\frac{\pi}{8}} \end{gathered}\begin{gathered} k=1\colon \\ z_1=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{1}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{1}{2}))) \\ z_1=\sqrt[4]{18}\cdot(\cos (\frac{5\pi}{8})+i\cdot\sin (\frac{5\pi}{8})) \\ z_1=\sqrt[4]{18}e^{i\frac{5\pi}{8}} \end{gathered}\begin{gathered} k=2\colon \\ z_2=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{2}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{2}{2}))) \\ z_2=\sqrt[4]{18}\cdot(\cos (\frac{9\pi}{8})+i\cdot\sin (\frac{9\pi}{8})) \\ z_2=\sqrt[4]{18}e^{i\frac{9\pi}{8}} \end{gathered}\begin{gathered} k=3\colon \\ z_3=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{3}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{3}{2}))) \\ z_3=\sqrt[4]{18}\cdot(\cos (\frac{13\pi}{8})+i\cdot\sin (\frac{13\pi}{8})) \\ z_3=\sqrt[4]{18}e^{i\frac{13\pi}{8}} \end{gathered}

Answer:

The four roots in exponential form are

z0 = 18^(1/4)*e^(i*π/8)

z1 = 18^(1/4)*e^(i*5π/8)

z2 = 18^(1/4)*e^(i*9π/8)

z3 = 18^(1/4)*e^(i*13π/8)

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