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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]

Sagot :

GinnyAnsweridn
Sure, let's determine the general form of the equation of a circle.

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

To convert this into the general form, we expand the left-hand side:

[tex]$(x - a)^2 + (y - b)^2 = m^2,$[/tex]
which expands to:
[tex]$x^2 - 2ax + a^2 + y^2 - 2by + b^2 = m^2.$[/tex]

Now, we bring all terms to one side of the equation to set it to zero:
[tex]$x^2 - 2ax + a^2 + y^2 - 2by + b^2 - m^2 = 0.$[/tex]

Rearranging the terms, we get:
[tex]$x^2 + y^2 - 2ax - 2by + (a^2 + b^2 - m^2) = 0.$[/tex]

Therefore, the general form of the equation of a circle with center at [tex]\((a, b)\)[/tex] and radius [tex]\(m\)[/tex] is:
[tex]$x^2 + y^2 - 2ax - 2by + (a^2 + b^2 - m^2) = 0.$[/tex]

Comparing this with the given options, the correct answer is:
A. [tex]\(x^2 + y^2 - 2ax - 2by + (a^2 + b^2 - m^2) = 0.\)[/tex]
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