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(x+5)^(2)+(y-6)^(2)=25 domain of the relation in interval notation.

User Imbr
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We are given the equation of a circle:

(x+5)^(2)+(y-6)^(2) = 25

From this equation, we can determine that the circle's center is at (-5, 6). This is because the form of a standard circle equation is (x-h)^2 + (y-k)^2 = r^2 where (h, k) is the center of the circle and r is the radius. Here, our standard circle equation is shifted by 5 units to the left and 6 units up, so our center is (-5, 6).

In addition, we can determine that the radius of the circle is 5 because r^2 = 25, so r = sqrt(25) = 5.

To find the domain of the relation, which is the set of possible x values, we need to look at the x-values at the most left and the most right points on the circle. These occur at the center of circle shifted left and right by the radius.

The leftmost point of the circle is at x = -5 - 5 and the rightmost point of the circle is at x = -5 + 5.

So, the domain of the relation is from -10 to 0.

In interval notation, this is expressed as [-10, 0].

User Vimal Bera
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