Question

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

Answer 1

The equation of the circle will be $x^2+(y-5)^2=25$

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

$(x-a)^2+(y-b)^2=r^2$ where:

r is the radius of the circle

(a, b) is the center

Given the centre (0, 5)

Get the radius

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

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

$(x-0)^2+(y-5)^2=5^2\\x^2+(y-5)^2=25$

This gives the equation of the circle.

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Answer 2

The equation of the circle is  $x^2 + (y-5)^2 = 25$

The graph of the circle is plotted below

The equation of a circle is of the form:

$(x - a)^2 + (y - b)^2 = r^2$

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

$Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance = \sqrt{(0-0)^2+(0-5)^2}\\Distance = \sqrt{25}\\Distance = 5$

The radius = 5

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

$(x - a)^2 + (y - b)^2 = r^2\\\\(x - 0)^2 + (y - 5)^2 = 5^2\\\\x^2 + (y-5)^2 = 25$

The equation of the circle is

$x^2 + (y-5)^2 = 25$

The graph of the circle is plotted below

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