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The equation of a circle is given below.

(x-20)^{2}+(y-0.05)^{2} = 81(x−20)

2

+(y−0.05)

2

=81left parenthesis, x, minus, 20, right parenthesis, squared, plus, left parenthesis, y, minus, 0, point, 05, right parenthesis, squared, equals, 81

What is its center?

((left parenthesis

,,comma

))right parenthesis

What is its radius?

If necessary, round your answer to two decimal places.

Respuesta :

abidemiokin

Answer:

Centre is (20, 0.05) and the radius is 9

Step-by-step explanation:

General form of 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 and (a, b) is the center of the circle.

Given the equation of a circle to be [tex](x-20)^{2}+(y-0.05)^{2} = 81[/tex], rewriting the equation to conform to the general equation above we have;

[tex](x-20)^{2}+(y-0.05)^{2} = 9^{2}[/tex]

Comparing the equation given to the general equation above, a = 20, b = 0.05 and the radius of the circle will be 9. The center (a, b) will be (20, 0.05).

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