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

• A. ${(x-4)}^2+{(y-6)}^2=7^2$; Center: $(4,6)$; Radius: $7$
• B. $\left(x-4\right)^2+\left(y-6\right)^2=7^2$; Center: $(6,4)$; Radius: $7$
• C. ${(x-6)}^2+{(y-4)}^2=7^2$; 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\( inside the parentheses for \)x$ terms.
4. Complete the square for the \(y\) terms: take half of \(-12\), which is \(-6\), then square it to get \$36\(. Add \)36\( inside the parentheses for \)y$ terms.
5. Since you added \$16\( and \)36\( to the left side, add \)16 + 36 = 52\( to the right side as well to keep the equation balanced. Then rewrite the equation as \)(x - 4)^2 + (y - 6)^2 = 3 + 52$.