Use the image to answer the question.

Ricardo draws a circle on the coordinate plane. The circle has the equation (16−(x+2)2)−−−−−−−−−−−−−√−5
(
16

(
𝑥
+
2
)
2
)

5
. The circle is now translated 3 units to the right and 3 units up. Where is the translated circle’s center?

(1 point)
(,)

1 answer

The center of the original circle can be found from its equation. The equation of the circle given is:

\[ y = \sqrt{16 - (x + 2)^2} - 5 \]

To write the equation in standard form, we can isolate \(y\):

\[ y + 5 = \sqrt{16 - (x + 2)^2} \]

Squaring both sides, we get:

\[ (y + 5)^2 = 16 - (x + 2)^2 \]

From \( (x + 2)^2 + (y + 5)^2 = 16 \), we can identify the center of the circle. The standard form of a circle's equation is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \((h, k)\) is the center and \(r\) is the radius.

From \((x + 2)^2\), we deduce that \(h = -2\).
From \((y + 5)^2\), we deduce that \(k = -5\).

Thus, the center of the original circle is:

\[ (-2, -5) \]

Now, when the circle is translated 3 units to the right, we add 3 to the \(x\)-coordinate of the center:

\[ -2 + 3 = 1 \]

When the circle is translated 3 units up, we add 3 to the \(y\)-coordinate of the center:

\[ -5 + 3 = -2 \]

Therefore, the new center of the translated circle is:

\[ (1, -2) \]

So the answer is:

\[ (1, -2) \]