{"id":2542,"date":"2024-08-05T22:20:38","date_gmt":"2024-08-05T22:20:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=2542"},"modified":"2025-08-15T16:35:35","modified_gmt":"2025-08-15T16:35:35","slug":"circles-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/circles-learn-it-2\/","title":{"raw":"Circles: Learn It 2","rendered":"Circles: Learn It 2"},"content":{"raw":"
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General Form of the Equation of a Circle<\/h2>\r\nIn algebra, circles can be described using different equations. One common way is the general form<\/strong>, which can look a bit complicated at first. This form doesn\u2019t directly show the center or the radius of the circle, making it harder to visualize and can be a little tricky to work with.\r\n\r\n
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general form of the equation of a circle<\/h3>\r\nThe general form of the equation of a circle is:\r\n

[latex]x^2+y^2+ax+by+c = 0[\/latex]<\/p>\r\n\r\n<\/section>To make things clearer, we often convert the general form to another version known as the standard form, <\/strong>[latex](x-h)^2+(y-k)^2 = r^2[\/latex]. This transformation involves a process called completing the square<\/strong>, in both [latex]x[\/latex] and [latex]y[\/latex], which helps us easily identify the circle\u2019s center and radius. Then, we can graph the circle using its center and radius.\r\n\r\n

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\r\n\r\nFind the center and radius, then graph the circle:\r\n

[latex]x^2+y^2-4x-6y+4 = 0[\/latex]<\/p>\r\n[reveal-answer q=\"726452\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"726452\"]\r\n

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\r\n\r\nWe need to rewrite this general form into standard form in order to find the center and radius.\r\n\r\nStep 1:<\/strong> Move the constant term to the other side\r\n

[latex]x^2 + y^2 - 4x - 6y = -4[\/latex]<\/p>\r\nStep 2:<\/strong> Group the [latex]x[\/latex] and [latex]y[\/latex] terms\r\n

[latex](x^2 - 4x) + (y^2 - 6y) = -4[\/latex]<\/p>\r\nStep 3:<\/strong> Complete the square for [latex]x[\/latex] and [latex]y[\/latex]\r\n

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  • For [latex]x[\/latex]: [latex]x^2 - 4x[\/latex]\r\nTake half of [latex]-4[\/latex], which is [latex]-2[\/latex], and square it, giving [latex]4[\/latex].\r\nAdd and subtract:<\/li>\r\n<\/ul>\r\n

    [latex]x^2 - 4x + 4 - 4 = (x^2 - 4x + 4) - 4 = (x - 2)^2 - 4[\/latex]<\/p>\r\n\r\n

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    • For [latex]y[\/latex]: [latex]y^2 - 6y[\/latex]\r\nTake half of [latex]-6[\/latex], which is [latex]-3[\/latex], and square it, giving [latex]9[\/latex].\r\nAdd and subtract:<\/li>\r\n<\/ul>\r\n

      [latex]y^2 - 6y + 9 - 9 = (y^2 - 6y + 9 )- 9 = (y- 3)^2 - 9[\/latex]<\/p>\r\nStep 4:<\/strong> Rewrite the equation with the completed squares\r\n

      [latex](x - 2)^2 - 4 + (y - 3)^2 - 9 = -4[\/latex]<\/p>\r\nStep 5:<\/strong> Simplify and move the constant terms to the other side\r\n

      [latex](x - 2)^2 + (y - 3)^2 - 13 = -4[\/latex]<\/p>\r\n

      [latex](x - 2)^2 + (y - 3)^2 = 9[\/latex]<\/p>\r\nNow we have the standard form: [latex](x - 2)^2 + (y - 3)^2 = 9[\/latex]\r\n\r\nThus, center: [latex](2,3)[\/latex] and radius: [latex]r = 3[\/latex].\r\n\r\nGraph:\r\n\r\n[caption id=\"attachment_2545\" align=\"aligncenter\" width=\"350\"]\"\" Circle on a coordinate plane with center and radius labeled[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>

      [ohm2_question hide_question_numbers=1]24883[\/ohm2_question]<\/section><\/section>
      [ohm2_question hide_question_numbers=1]24878[\/ohm2_question]<\/section>
      [ohm2_question hide_question_numbers=1]24877[\/ohm2_question]<\/section><\/div>\r\n<\/div>\r\n<\/div>","rendered":"
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      General Form of the Equation of a Circle<\/h2>\n

      In algebra, circles can be described using different equations. One common way is the general form<\/strong>, which can look a bit complicated at first. This form doesn\u2019t directly show the center or the radius of the circle, making it harder to visualize and can be a little tricky to work with.<\/p>\n

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      general form of the equation of a circle<\/h3>\n

      The general form of the equation of a circle is:<\/p>\n

      [latex]x^2+y^2+ax+by+c = 0[\/latex]<\/p>\n<\/section>\n

      To make things clearer, we often convert the general form to another version known as the standard form, <\/strong>[latex](x-h)^2+(y-k)^2 = r^2[\/latex]. This transformation involves a process called completing the square<\/strong>, in both [latex]x[\/latex] and [latex]y[\/latex], which helps us easily identify the circle\u2019s center and radius. Then, we can graph the circle using its center and radius.<\/p>\n

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      Find the center and radius, then graph the circle:<\/p>\n

      [latex]x^2+y^2-4x-6y+4 = 0[\/latex]<\/p>\n