reduce the general equation of the circle into standard form 5x²2+5y²+10x-5y+3=0​

Answer:To reduce the general equation of a circle into its standard form, follow these steps:Given equation:\[ 5x^2 + 5y^2 + 10x - 5y + 3 = 0 \]**Step 1: Divide the entire equation by 5 to simplify:**\[ x^2 + y^2 + 2x - y + \frac{3}{5} = 0 \]**Step 2: Rearrange the terms to prepare for completing the square:**\[ x^2 + 2x + y^2 - y = -\frac{3}{5} \]**Step 3: Complete the square for \(x\) and \(y\).**For \(x\):\[ x^2 + 2x \]To complete the square, add and subtract \((\frac{2}{2})^2 = 1\):\[ x^2 + 2x + 1 - 1 \]\[ (x + 1)^2 - 1 \]For \(y\):\[ y^2 - y \]To complete the square, add and subtract \((\frac{-1}{2})^2 = \frac{1}{4}\):\[ y^2 - y + \frac{1}{4} - \frac{1}{4} \]\[ \left(y - \frac{1}{2}\right)^2 - \frac{1}{4} \]**Step 4: Substitute the completed squares back into the equation:**\[ (x + 1)^2 - 1 + \left(y - \frac{1}{2}\right)^2 - \frac{1}{4} = -\frac{3}{5} \]**Step 5: Combine and simplify the constants:**\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 - 1 - \frac{1}{4} = -\frac{3}{5} \]\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 - \frac{5}{4} = -\frac{3}{5} \]\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 = -\frac{3}{5} + \frac{5}{4} \]Convert to a common denominator (20):\[ -\frac{3}{5} = -\frac{12}{20} \]\[ \frac{5}{4} = \frac{25}{20} \]\[ -\frac{12}{20} + \frac{25}{20} = \frac{13}{20} \]So the standard form of the circle is:\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{13}{20} \]This is the standard form of the circle, with center at \((-1, \frac{1}{2})\) and radius \(\sqrt{\frac{13}{20}}\).

Answered by MrProject | 2024-09-01