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

${(x - 4)}^{2} + {(y - 6)}^{2} = 7^{2}$ ; Center: $open paren 4 comma 6 close paren$ ; Radius: $7$

$$\left$(x-4$\right$)^2+$\left$(y-6$\right$)^2=7^2$ ; Center: $open paren 6 comma 4 close paren$ ; Radius: $7$

${(x - 6)}^{2} + {(y - 4)}^{2} = 7^{2}$ ; Center: $$\left$(6,4$\right$)$ ; Radius: $7$

$$\left$(x-4$\right$)^2+$\left$(y-6$\right$)^2=7^2$ ; Center: $$\left$(4,6$\right$)$ ; Radius: $49$

Verified step by step guidance

Start with the given equation of the circle: $x^2 + y^2 - 8x - 12y + 3 = 0$.

Group the $x$ terms and $y$ terms together: $(x^2 - 8x) + (y^2 - 12y) = -3$ (move the constant to the right side).

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.

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.

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 $(x - 4)^2 + (y - 6)^2 = 3 + 52$.

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Write the standard form equation of the circle described. Give the center and radius. x2+y2−8x−12y+3=0x^2+y^2-8x-12y+3=0

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