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Find an equation of the circle with the following characteristics and sketch its graph a center (0, 5), contains (0, 0)

Sagot :

abidemiokin

The equation of the circle will be [tex]x^2+(y-5)^2=25[/tex]

The formula finding the equation of a circle is expressed as:

[tex](x-a)^2+(y-b)^2=r^2[/tex] where:

r is the radius of the circle

(a, b) is the center

Given the centre (0, 5)

Get the radius

[tex]r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\r=\sqrt{(0-0)^2+(0-5)^2}\\r=\sqrt{25}\\r=5units[/tex]

Substitute the radius  and the centre into the equation of a circle as shown:

[tex](x-0)^2+(y-5)^2=5^2\\x^2+(y-5)^2=25[/tex]

This gives the equation of the circle.

Learn more here: https://brainly.com/question/24217736

View image abidemiokin

The equation of the circle is  [tex]x^2 + (y-5)^2 = 25[/tex]

The graph of the circle is plotted below

The equation of a circle is of the form:

[tex](x - a)^2 + (y - b)^2 = r^2[/tex]

where (a, b) is the center of the circle

In this question:

The center, (a, b)  =  (0, 5)

The radius is the distance between (0, 5) and (0, 0)

Find the distance using the formula below

[tex]Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance = \sqrt{(0-0)^2+(0-5)^2}\\Distance = \sqrt{25}\\Distance = 5[/tex]

The radius = 5

Substitute a = 0. b = 5, and r = 5 into the equation of a circle

[tex](x - a)^2 + (y - b)^2 = r^2\\\\(x - 0)^2 + (y - 5)^2 = 5^2\\\\x^2 + (y-5)^2 = 25\\[/tex]

The equation of the circle is

[tex]x^2 + (y-5)^2 = 25[/tex]

The graph of the circle is plotted below

Learn more here: https://brainly.com/question/23226948

View image Adetunmbiadekunle
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