Answered

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1a. What is the center of the circle given by the equation [tex]x^2 + y^2 = 64[/tex]?

A. [tex]\((6, 4)\)[/tex]
B. [tex]\((64, 0)\)[/tex]
C. [tex]\((0, -64)\)[/tex]
D. [tex]\((0, 64)\)[/tex]
E. [tex]\((0, 0)\)[/tex]

Sagot :

GinnyAnswer
To find the center of the circle given by the equation [tex]\( x^2 + y^2 = 64 \)[/tex], we need to recognize the standard form of a circle's equation. The general form of a circle's equation is:

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

Here, [tex]\((h, k)\)[/tex] represents the center of the circle, and [tex]\(r\)[/tex] is the radius.

In the given equation [tex]\( x^2 + y^2 = 64 \)[/tex], we can compare it to the general form:

1. [tex]\( (x - 0)^2 + (y - 0)^2 = 64 \)[/tex]

From this comparison, we see that [tex]\( h = 0 \)[/tex] and [tex]\( k = 0 \)[/tex]. Therefore, the center of the circle is at [tex]\((0,0)\)[/tex].

So, the answer is:

[tex]\[ \boxed{(0,0)} \][/tex]
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