Answer :
Sure, let's write the standard equation of a circle with the given center [tex]\((-4, -1)\)[/tex] and radius 12 units.
The standard form of the equation of a circle with center [tex]\((h, k)\)[/tex] and radius [tex]\(r\)[/tex] is given by:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
In this case, the center [tex]\((h, k)\)[/tex] is [tex]\((-4, -1)\)[/tex], and the radius [tex]\(r\)[/tex] is 12 units. Substituting these values into the standard form equation, we get:
[tex]\[ (x - (-4))^2 + (y - (-1))^2 = 12^2 \][/tex]
Simplifying the equation:
[tex]\[ (x + 4)^2 + (y + 1)^2 = 144 \][/tex]
So, the standard equation of the circle with center [tex]\((-4, -1)\)[/tex] and radius 12 units is:
[tex]\[ (x + 4)^2 + (y + 1)^2 = 144 \][/tex]
The standard form of the equation of a circle with center [tex]\((h, k)\)[/tex] and radius [tex]\(r\)[/tex] is given by:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
In this case, the center [tex]\((h, k)\)[/tex] is [tex]\((-4, -1)\)[/tex], and the radius [tex]\(r\)[/tex] is 12 units. Substituting these values into the standard form equation, we get:
[tex]\[ (x - (-4))^2 + (y - (-1))^2 = 12^2 \][/tex]
Simplifying the equation:
[tex]\[ (x + 4)^2 + (y + 1)^2 = 144 \][/tex]
So, the standard equation of the circle with center [tex]\((-4, -1)\)[/tex] and radius 12 units is:
[tex]\[ (x + 4)^2 + (y + 1)^2 = 144 \][/tex]
