"3x^2+3y^2-6x-8y-12 = 3x^2 - 6x + 3 - 3 +3y^2 - 8y + \\frac{16}{3} - \\frac{16}{3} - 12 = \\\\ =(\\sqrt{3}x-\\sqrt{3})^2 + (\\sqrt{y}- \\frac{4}{\\sqrt{3}})^2 - \\frac{61}{3}\n\\\\ \u00a0\\\\\u00a0\\\\\n(\\sqrt{3}x-\\sqrt{3})^2 + (\\sqrt{y}- \\frac{4}{\\sqrt{3}})^2 = \\frac{61}{3} \\\\ This\u00a0is\u00a0circle"
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