Answer :
To determine the equation of a circle given its center and diameter, follow these steps:
1. Identify the center of the circle, which is given as [tex]\((-5, 2)\)[/tex]. This means [tex]\(h = -5\)[/tex] and [tex]\(k = 2\)[/tex].
2. Calculate the radius of the circle. The diameter is given as 12, so the radius [tex]\(r\)[/tex] is half of that:
[tex]\[ r = \frac{12}{2} = 6 \][/tex]
3. Square the radius to use it in the standard form of the equation of a circle:
[tex]\[ r^2 = 6^2 = 36 \][/tex]
4. Substitute the center coordinates [tex]\((h, k)\)[/tex] and the radius squared [tex]\(r^2\)[/tex] into the standard form of the equation of a circle:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
Substituting [tex]\(h = -5\)[/tex], [tex]\(k = 2\)[/tex], and [tex]\(r^2 = 36\)[/tex], we get:
[tex]\[ (x - (-5))^2 + (y - 2)^2 = 36 \][/tex]
Simplifying, we have:
[tex]\[ (x + 5)^2 + (y - 2)^2 = 36 \][/tex]
Therefore, the correct answer in standard form for the given circle is:
[tex]\[ (x + 5)^2 + (y - 2)^2 = 36 \][/tex]
Thus, the correct answer to the question is:
[tex]\[ \boxed{(x+5)^2+(y-2)^2=36} \][/tex]
1. Identify the center of the circle, which is given as [tex]\((-5, 2)\)[/tex]. This means [tex]\(h = -5\)[/tex] and [tex]\(k = 2\)[/tex].
2. Calculate the radius of the circle. The diameter is given as 12, so the radius [tex]\(r\)[/tex] is half of that:
[tex]\[ r = \frac{12}{2} = 6 \][/tex]
3. Square the radius to use it in the standard form of the equation of a circle:
[tex]\[ r^2 = 6^2 = 36 \][/tex]
4. Substitute the center coordinates [tex]\((h, k)\)[/tex] and the radius squared [tex]\(r^2\)[/tex] into the standard form of the equation of a circle:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
Substituting [tex]\(h = -5\)[/tex], [tex]\(k = 2\)[/tex], and [tex]\(r^2 = 36\)[/tex], we get:
[tex]\[ (x - (-5))^2 + (y - 2)^2 = 36 \][/tex]
Simplifying, we have:
[tex]\[ (x + 5)^2 + (y - 2)^2 = 36 \][/tex]
Therefore, the correct answer in standard form for the given circle is:
[tex]\[ (x + 5)^2 + (y - 2)^2 = 36 \][/tex]
Thus, the correct answer to the question is:
[tex]\[ \boxed{(x+5)^2+(y-2)^2=36} \][/tex]