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Write the standard equation of the circle below.



(x + 1)2 + (y - 2)^2 = 1

(x - 1)2 + (y - 2)^2 = 1

(x + 1)2 + (y + 2)^2 = 1

(x - 1)2 + (y - 2)^2 = 1

Write the standard equation of the circle below x 12 y 22 1 x 12 y 22 1 x 12 y 22 1 x 12 y 22 1 class=

Respuesta :

calculista

Answer:

[tex](x+1)^{2} +(y-2)^{2} =1[/tex]

Step-by-step explanation:

we know that

the equation of the circle into center radius form is equal to

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

where

r is the radius of the circle

(h,k) is the center of the circle

In this problem we have

[tex](h,k)=(-1,2)[/tex]

[tex]r=1\ units[/tex]

substitute

[tex](x+1)^{2} +(y-2)^{2} =1^{2}[/tex]

[tex](x+1)^{2} +(y-2)^{2} =1[/tex]

zainebamir540

Answer:

Choice  1 is correct answer.

Step-by-step explanation:

Since , we know that

(x-a)²+(y-b)² = r² is equation of circle Where r is radius of circle and (a,b) is center of circle.

From diagram , we notice that

center of circle is (a,b) = (-1,2)

Radius is 1.

a = -1 and b =2   and r = 1

Putting above values in equation of circle, we have

(x-(-1))²+(y-(2))² = (1)²

simplifying

(x+1)²+(y-2)² = 1   which is the answer.

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