Write the standard form of the equation of the circle with center (-5, 2) and radius 4. What is the domain and range?

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The standard equation of a circle with center (h, k) and radius r is (x - h)^2 + (y - k)^2 = r^2

Here the center is (-5, 2) and the radius is 4.

The standard equation of the circle is (x + 5)^2 + (y - 2)^2 = 4^2

To find the domain and range we take a function, the graph of which includes the circle derived above and all points that lie within it.

The values that x can take for which y is defined lie in the interval [-9, -1]. The domain is therefore [-9, -1].

The range is all the values of y that result from values of x that lie in the domain. The range is given by [-2, 6].

We'll recall what is the standard form of the equation of circle:

(x-h)^2 + (y-k)^2 = r^2, where the coordinates of the center of circle are represented by the pair (h,k) and r is the radius of the circle:

The standard form of the equation of circle is:

(x +5 )^2 + (y - 2)^2 = 4^2

The domain of the definition consists of all x values that makes the equation possible:

(x +5 )^2 = 4^2

We'll take square root both sides:

x + 5 = 4 => x = -1

x + 5 = -4 => x = -9

The domain of definition of the function is [-9 ; -1].

The range of the values of function is:

(y - 2)^2 = 4^2

y - 2 = 4 => y = 6

y - 2 = -4 => y = -2

The range of the values of function is [-2 ; 6].

**Therefore, the requested standard form of the equation of circle andthe domain and the range of the function are: (x +5 )^2 + (y - 2)^2 = 4^2 ; domain [-9 ; -1] and the range [-2 ; 6].**

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