Write the equation in standard form for the circle $x^2+y^2-4x+7= 0$ .

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Algebra 2

Convert equations of circles from general to standard form

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Q. Write the equation in standard form for the circle

x^2+y^2-4x+7= 0

Identify equation and need: Identify the given equation and the need to complete the square for $x$ . $\newline$ The given equation is $x^2 + y^2 - 4x + 7 = 0$ . To get it into standard form, we need to complete the square for the $x$ terms.
Group and leave space: Group the $x$ terms and leave a space for completing the square. $x^2 - 4x + (\ ) + y^2 = -7$ We will add a certain number inside the parentheses to complete the square for the $x$ terms.
Calculate number to complete: Calculate the number needed to complete the square for the x terms. $\newline$ To complete the square, we take the coefficient of x, which is $-4$ , divide it by $2$ , and square it. $(-4/2)^2 = 4$ . $\newline$ Add $4$ inside the parentheses to complete the square for the x terms.
Add number to both sides: Add the same number to both sides of the equation to maintain equality. $\newline$ $x^2 - 4x + 4 + y^2 = -7 + 4$ $\newline$ Now the equation is balanced.
Factor and simplify: Factor the completed square for the x terms and simplify the right side of the equation. $\newline$ $(x - 2)^2 + y^2 = -3$ $\newline$ This is the equation with the x terms factored.
Ensure positive right side: Since the standard form of a circle's equation is $(x - h)^2 + (y - k)^2 = r^2$ , we need to ensure the right side of the equation is positive because the radius squared must be a positive number. $\newline$ However, we have a negative number on the right side, which indicates a math error. A circle cannot have a negative radius squared.