Question

rewrite the equation of the circle (x^{2}+y^{2}-4x + 6y-3 = 0) into standard form. enter your answers in the boxes to complete the equation. be sure to include an action or situation symbol where needed. ((xsquare)^{2}+(ysquare)^{2}=square)

Explanation:

Step1: Complete the square for x - terms

The x - terms are \(x^{2}-4x\). Using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), for \(x^{2}-4x\), we have \(a = x\) and \(2b = 4\) (so \(b = 2\)), and \(x^{2}-4x=(x - 2)^{2}-4\).

Step2: Complete the square for y - terms

The y - terms are \(y^{2}+6y\). Using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), for \(y^{2}+6y\), we have \(a = y\) and \(2b=6\) (so \(b = 3\)), and \(y^{2}+6y=(y + 3)^{2}-9\).

Step3: Rewrite the original equation

The original equation \(x^{2}+y^{2}-4x + 6y-3=0\) can be rewritten as \((x^{2}-4x)+(y^{2}+6y)=3\). Substituting the completed - square expressions, we get \((x - 2)^{2}-4+(y + 3)^{2}-9 = 3\).

Step4: Simplify to standard form

\((x - 2)^{2}+(y + 3)^{2}=3 + 4+9\).

Answer:

\((x - 2)^{2}+(y + 3)^{2}=16\)