Select the correct answer.

What is the general form of the equation of a circle with center at [tex]\((a, b)\)[/tex] and radius of length [tex]\(m\)[/tex]?

A. [tex]\(x^2 + y^2 - 2ax - 2by + (a^2 + b^2 - m^2) = 0\)[/tex]

B. [tex]\(x^2 + y^2 + 2ax + 2by + (a^2 + b^2 - m^2) = 0\)[/tex]

C. [tex]\(x^2 + y^2 - 2ax - 2by + (a + b - m^2) = 0\)[/tex]

D. [tex]\(x^2 + y^2 + 2ax + 2by + a^2 + b^2 = -m^2\)[/tex]



Answer :

To determine the correct general form of the equation of a circle with a given center [tex]\((a, b)\)[/tex] and a radius [tex]\(m\)[/tex], let's begin by recalling the standard circle equation and then derivatively transform it into its general form.

1. Standard Equation:
The standard form of the equation of a circle with center at [tex]\((a, b)\)[/tex] and radius [tex]\(m\)[/tex] is given by:
[tex]\[ (x - a)^2 + (y - b)^2 = m^2 \][/tex]

2. Expand the Equation:
We will expand the binomials in the equation:
[tex]\[ (x - a)^2 + (y - b)^2 = m^2 \][/tex]
Expanding the square terms, we get:
[tex]\[ (x^2 - 2ax + a^2) + (y^2 - 2by + b^2) = m^2 \][/tex]
Combine like terms:
[tex]\[ x^2 - 2ax + a^2 + y^2 - 2by + b^2 = m^2 \][/tex]

3. Rearrange to General Form:
Move all terms to one side of the equation to get the general form:
[tex]\[ x^2 + y^2 - 2ax - 2by + a^2 + b^2 - m^2 = 0 \][/tex]

Now let’s match this derived equation with the given options:

- Option A: [tex]\(x^2 + y^2 - 2ax - 2by + (a^2 + b^2 - m^2) = 0\)[/tex]
- This matches our derived equation exactly.

- Option B: [tex]\(x^2 + y^2 + 2ax + 2by + (a^2 + b^2 - m^2) = 0\)[/tex]
- This has the wrong signs in front of [tex]\(2ax\)[/tex] and [tex]\(2by\)[/tex].

- Option C: [tex]\(x^2 + y^2 - 2ax - 2by + (a + b - m^2) = 0\)[/tex]
- This has an incorrect term [tex]\((a + b - m^2)\)[/tex] instead of [tex]\((a^2 + b^2 - m^2)\)[/tex].

- Option D: [tex]\(x^2 + y^2 + 2ax + 2by + a^2 + b^2 = -m^2\)[/tex]
- This has incorrect signs and incorrectly arranged terms.

Therefore, the correct answer is:

Option A: [tex]\(x^2 + y^2 - 2ax - 2by + (a^2 + b^2 - m^2) = 0\)[/tex]