Determine the equation of the circle with center (0, -5) containing the point
(-√109,-6).

Answers

Answer 1

x^2 + y^2 - 10y + (-85) is equation of the circle with center (0, -5) containing the point (-√109,-6).

How do you find the equation of a circle?

To find the equation of a circle with a given center and containing a given point, we can use the distance formula. The distance formula is given by:

d = √((x2 - x1)^2 + (y2 - y1)^2)

where (x1, y1) is the center of the circle and (x2, y2) is the point on the circle. In this case, the center of the circle is (0, -5) and the point on the circle is (-√109, -6). Plugging these values into the distance formula, we get:

d = √((-√109 - 0)^2 + (-6 - (-5))^2)

d = √((-√109)^2 + (-1)^2)

d = √(109 + 1)

d = √110

Since the point (-√109, -6) is on the circle, the distance from the center of the circle to this point is the radius of the circle. Therefore, the radius of the circle is √110.

The equation of a circle with center (h, k) and radius r is given by:

(x - h)^2 + (y - k)^2 = r^2

In this case, the center of the circle is (0, -5) and the radius is √110, so the equation of the circle is:

(x - 0)^2 + (y - (-5))^2 = (√110)^2

x^2 + y^2 - 10y + 25 = 110

x^2 + y^2 - 10y + (-85) = 0

This is the equation of the circle with center (0, -5) and containing the point (-√109, -6).

To learn more about circle equation refer :

https://brainly.com/question/1506955

#SPJ1


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I believe the answer to 7 would be;
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Another train of thought was the number in the middle of 6 and 10 is 8 and 8(8) = 64
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Answers

Answer:

(i) The estimated regression equation is;

[tex]\hat y[/tex] ≈ 1.6896 + 0.0604·X

The coefficient of 'X' indicates that [tex]\hat y[/tex] increase by a multiple of 0.0604 for each million dollar increase in sales, X

(ii) The estimated earnings for the company is approximately $4.7096 million

(iii) The standard error of estimate is approximately 29.34

The high standard error of estimate indicates that individual mean do not accurately represent the population mean

(iv) The coefficient of determination is approximately 0.57925

The coefficient of determination indicates that the probability of the coordinate of a new point of data to be located on the line is 0.57925

Step-by-step explanation:

The given data is presented as follows;

[tex]\begin{array}{ccc}Sales \ (\$million)&&Earning \ (\$million) \\89.2&&4.9\\18.6&&4.4\\18.2&&1.3\\71.7&&8\\58.6&&6.6\\46.8&&4.1\\17.5&&2.6\\11.9&&1.7\end{array}[/tex]

(i) From the data, we have;

The regression equation can be presented as follows;

[tex]\hat y[/tex] = b₀ + b₁·x

Where;

b₁ = The slope given as follows;

[tex]b_1 = \dfrac{\Sigma(x_i - \overline x) \cdot (y_i - \overline y)}{\Sigma(x_i - \overline x)^2}[/tex]

b₀ = [tex]\overline y[/tex] - b₁·[tex]\overline x[/tex]

From the data, we have;

[tex]{\Sigma(x_i - \overline x) \cdot (y_i - \overline y)}[/tex] = 364.05

[tex]\Sigma(x_i - \overline x)^2}[/tex] = 6,027.259

[tex]\overline y[/tex] = 4.2

[tex]\overline x[/tex] = 41.5625

∴ b₁ = 364.05/6,027.259 ≈ 0.06040059005

b₀ = 4.2 - 0.06040059005 × 41.5625 ≈ 1.68960047605 ≈ 1.69

Therefore, we have the regression equation as follows;

[tex]\hat y[/tex] ≈ 1.6896 + 0.0604·X

The coefficient of 'X' indicates that the earnings increase by a multiple of 0.0604 for each million dollar increase in sales

(ii) For the small company, we have;

X = $50.0 million, therefore, we get;

[tex]\hat y[/tex] = 1.6896 + 0.0604 × 50 = 4.7096

The estimated earnings for the company, [tex]\hat y[/tex] = 4.7096 million

(iii) The standard error of estimate, σ, is given by the following formula;

[tex]\sigma =\sqrt{\dfrac{\sum \left (x_i-\mu \right )^{2} }{n - 1}}[/tex]

Where;

n = The sample size

Therefore, we have;

[tex]\sigma =\sqrt{\dfrac{6,027.259 }{8 - 1}} \approx 29.34[/tex]

The standard error of estimate, σ ≈ 29.34

The high standard error of estimate indicates that it is very unlikely that a given mean value within the data is a representation of the true population mean

(iv) The coefficient of determination (R Square) is given as follows;

[tex]R^2 = \dfrac{SSR}{SST}[/tex]

Where;

SSR = The Sum of Squared Regression ≈ 21.9884

SST = The total variation in the sample ≈ 37.96

Therefore, R² ≈ 21.9884/37.96 ≈ 0.57925

The coefficient of determination, R² ≈ 0.57925.

Therefore, by the coefficient of determination, the likelihood of a new introduced data point to located on the line is 0.57925

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