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The center of a circle is located at \((3, 8)\), and the circle has a radius that is 5 units long. What is the general form of the equation for the circle?

A. \(x^2 + y^2 - 6x - 16y + 48 = 0\)
B. \(x^2 + y^2 - 6x - 16y - 25 = 0\)
C. \(x^2 + y^2 + 6x + 16y + 48 = 0\)
D. [tex]\(x^2 + y^2 + 6x + 16y - 25 = 0\)[/tex]


Sagot :

To find the equation of a circle given its center and radius, we start with the standard form of a circle's equation, which is:

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

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

Given:
- The center of the circle \((h, k) = (3, 8)\)
- The radius \(r = 5\)

We substitute these values into the standard form:

[tex]\[ (x - 3)^2 + (y - 8)^2 = 5^2 \][/tex]

Simplify the radius squared:

[tex]\[ (x - 3)^2 + (y - 8)^2 = 25 \][/tex]

Next, we expand the squared binomials:

[tex]\[ (x - 3)^2 = x^2 - 6x + 9 \][/tex]

[tex]\[ (y - 8)^2 = y^2 - 16y + 64 \][/tex]

Putting it all together:

[tex]\[ x^2 - 6x + 9 + y^2 - 16y + 64 = 25 \][/tex]

Combine all the terms on the left side to form a single equation, moving the constant from the right side to the left:

[tex]\[ x^2 + y^2 - 6x - 16y + 9 + 64 - 25 = 0 \][/tex]

Simplify the constants:

[tex]\[ x^2 + y^2 - 6x - 16y + 48 = 0 \][/tex]

Thus, the general form of the equation for this circle is:

[tex]\[ x^2 + y^2 - 6x - 16y + 48 = 0 \][/tex]

Therefore, the correct answer is:
A. [tex]\(x^2 + y^2 - 6 x - 16 y + 48 = 0\)[/tex]