Find the equation of a circle whose center is [tex]\((-4,11)\)[/tex] and radius is [tex]\(\sqrt{5}\)[/tex] units.

A. [tex]\(x^2 + y^2 + 8x + 22y + 132 = 0\)[/tex]
B. [tex]\(x^2 + y^2 + 8x - 22y + 132 = 0\)[/tex]
C. [tex]\(x^2 + y^2 + 8x + 22y - 135 = 0\)[/tex]
D. [tex]\(x^2 - y^2 + 8x + 22y - 135 = 0\)[/tex]



Answer :

To find the equation of a circle, you can use the standard form:

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

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

Given the center is [tex]\((-4, 11)\)[/tex] and the radius is [tex]\(\sqrt{5}\)[/tex], the standard form of the equation of the circle is:

[tex]\[(x + 4)^2 + (y - 11)^2 = (\sqrt{5})^2\][/tex]

This can be simplified to:

[tex]\[(x + 4)^2 + (y - 11)^2 = 5\][/tex]

Now let's expand this equation:

[tex]\[ (x + 4)^2 + (y - 11)^2 = 5 \][/tex]
[tex]\[ = (x^2 + 8x + 16) + (y^2 - 22y + 121) = 5 \][/tex]

Combine the terms:

[tex]\[ x^2 + 8x + 16 + y^2 - 22y + 121 = 5 \][/tex]

Simplify by moving all terms to one side of the equation:

[tex]\[ x^2 + 8x + y^2 - 22y + 137 = 5 \][/tex]

[tex]\[ x^2 + 8x + y^2 - 22y + 132 = 0 \][/tex]

So, the final equation of the circle in its general form is:

[tex]\[ x^2 + y^2 + 8x - 22y + 132 = 0 \][/tex]

Therefore, the correct equation is:

[tex]\[ x^2 + y^2 + 8x - 22y + 132 = 0 \][/tex]