Question

A circle is drawn on a coordinate plane with the center point at (5,−3) and a radius of 6 . Derive the equation of the circle in standard form, using the given information and the Pythagorean Theorem.

(1 point)
Responses

(x+5)2+(y−3)2=36open paren x plus 5 close paren squared plus open paren y minus 3 close paren squared is equal to 36

(x+3)2+(y−5)2=36 open paren x plus 3 close paren squared plus open paren y minus 5 close paren squared is equal to 36

(x−5)2+(y+3)2=36 open paren x minus 5 close paren squared plus open paren y plus 3 close paren squared is equal to 36

(x−3)2+(y−5)2=36

Answer 1

To write the equation of a circle in standard form, we use the formula:

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

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

Given:
- Center \((h, k) = (5, -3)\)
- Radius \(r = 6\)

First, we need to compute \(r^2\):
\[
r^2 = 6^2 = 36
\]

Now we can substitute \(h\), \(k\), and \(r^2\) into the standard form equation:

\[
(x - 5)^2 + (y - (-3))^2 = 36
\]

This simplifies to:

\[
(x - 5)^2 + (y + 3)^2 = 36
\]

So, the equation of the circle in standard form is:

\[
(x - 5)^2 + (y + 3)^2 = 36
\]

From the options provided, the correct response is:

**(x−5)²+(y+3)²=36** (open paren x minus 5 close paren squared plus open paren y plus 3 close paren squared is equal to 36)