Write the standard form equation of the circle described. Give the center and radius.
$x^2+y^2-8x-12y+3=0$

• A. $\left(x-4\right)^2+\left(y-6\right)^2=7^2$; Center: $\left(4,6\right)$; Radius: $7$
• B. $\left(x-4\right)^2+\left(y-6\right)^2=7^2$; Center: $\left(6,4\right)$; Radius: $7$
• C. $\left(x-6\right)^2+\left(y-4\right)^2=7^2$; Center: $\left(6,4\right)$; Radius: $7$
• D. $\left(x-4\right)^2+\left(y-6\right)^2=7^2$; Center: $\left(4,6\right)$; Radius: $49$

Answer: (A)

Verified step by step guidance

1. Start with the given equation of the circle: \(x^2 + y^2 - 8x - 12y + 3 = 0\).
2. Group the \(x\) terms and \(y\) terms together: \((x^2 - 8x) + (y^2 - 12y) = -3\) (move the constant to the right side).
3. Complete the square for the \(x\) terms: take half of \(-8\), which is \(-4\), then square it to get \$16\(. Add \)16\( inside the \)x\( group and also add \)16$ to the right side to keep the equation balanced.
4. Complete the square for the \(y\) terms: take half of \(-12\), which is \(-6\), then square it to get \$36\(. Add \)36\( inside the \)y\( group and also add \)36$ to the right side.
5. Rewrite the equation in standard form: \((x - 4)^2 + (y - 6)^2 = \) (sum of the numbers on the right side). From this, identify the center as \((4, 6)\) and the radius as the square root of the right side.