IDNLearn.com is your trusted platform for finding reliable answers. Find reliable solutions to your questions quickly and accurately with help from our dedicated community of experts.

Identify the radius and the center of a circle whose equation is [tex](x-5)^2 + y^2 = 81[/tex].

The radius of the circle is [tex]\square[/tex] units.

The center of the circle is at ([tex]\square[/tex], [tex]\square[/tex]).


Sagot :

Sure, let's solve this step-by-step.

We start with the equation of the circle given in the standard form:

[tex]\[ (x - 5)^2 + y^2 = 81 \][/tex]

The standard form of the equation of a circle is:

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

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

By comparing the given equation [tex]\((x - 5)^2 + y^2 = 81\)[/tex] with the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], we can identify the following:

1. The term [tex]\((x - 5)^2\)[/tex] tells us that [tex]\(h = 5\)[/tex].
2. The term [tex]\(y^2\)[/tex] can be rewritten as [tex]\((y - 0)^2\)[/tex], which tells us that [tex]\(k = 0\)[/tex].
3. The right-hand side of the equation is [tex]\(81\)[/tex], which equals [tex]\(r^2\)[/tex]. Therefore, [tex]\(r^2 = 81\)[/tex].

To find the radius [tex]\(r\)[/tex], we take the square root of both sides:

[tex]\[ r = \sqrt{81} \][/tex]

Solving this, we get:

[tex]\[ r = 9 \][/tex]

Therefore, the radius of the circle is 9 units, and the center of the circle is at [tex]\((5, 0)\)[/tex].

Putting this information into the given format:

The radius of the circle is [tex]\(9\)[/tex] units.
The center of the circle is at [tex]\((5, 0)\)[/tex].