# What is the center and radius of the circle: x^2 + y^2 – 4x + 8y - 5 = 0. Also, what is the domain and range of the function?

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### 2 Answers

The equation of the circle given is x^2 + y^2 – 4x + 8y - 5 = 0

x^2 + y^2 – 4x + 8y - 5 = 0

completing the squares

x^2 - 4x + 4 + y^2 + 8y + 16 = 5 + 4 + 16

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

This is in the standard form with center (2, -4) and radius 5.

Consider a function that includes the circle and the points that lie within it: (y + 4)^2 = 5^2 - (x - 2)^2

The domain of the function is all the values that x can take for y to have a real value. The domain is the set of values [-3, 7]

The range of the function is the values that y takes for values of x lying in the domain. The range is the set of values [-9, 1]

Regroup the original equationto be

(x^2 - 4x) + (y^2 + 8y) = 5

the form of a binomial (a+b)^2 = a^2 + 2ab + b^2 so the middle term is of the form 2ab.

so 2ab = -4 where a = 1 so b = -2 and b^2 = 4. Likewise 2ab = 8 where a=1 so b = 4 and b^2 = 16

Our equations no look like

(x^2 - 4x + 4) +(y^2 +8y + 16) = 5 + 4 + 16

so (x-2)^2 + (y + 4)^2 = 25

Our center is at (2,-4) with a radius of 5

the Domain is [-3, 7] and the range is [1,-9]