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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, start superscript, 2, end superscript, plus, left parenthesis, y, minus, 0, point, 05, right parenthesis, start superscript, 2, end superscript, 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 :

calculista

Answer:

Part 1) The center is the point [tex](20,0.05)[/tex]

Part 2) The radius is equal to [tex]r=9\ units[/tex]

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

[tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex]

where

(h,k) is the center of the circle

r is the radius of the circle

In this problem we have

[tex](x-20)^{2}+(y-0.05)^{2}=81[/tex]

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

so

[tex](h,k)=(20,0.05)[/tex]

[tex]r=9\ units[/tex]

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