{"id":3442,"date":"2021-07-24T13:14:49","date_gmt":"2021-07-24T13:14:49","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3442"},"modified":"2021-11-20T22:09:57","modified_gmt":"2021-11-20T16:39:57","slug":"diameter-form-of-circle","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/diameter-form-of-circle\/","title":{"rendered":"Diameter Form of Circle – Equation and Examples"},"content":{"rendered":"
The equation of the circle drawn on the straight line joining two points \\((x_1, y_1)\\) and \\((x_2, y_2)\\) is<\/p>\n
\n(\\(x-x_1\\))(\\(x-x_2\\)) + (\\(y-y_1\\))(\\(y-y_2\\)) = 0<\/strong>.<\/p>\n<\/blockquote>\n
This is known as diameter form of circle<\/strong>.<\/p>\n
where \\((x_1, y_1)\\) and \\((x_2, y_2)\\)<\/strong> are coordinates of two ends of diameter of circle.<\/p>\n
Note<\/strong> – If the coordinates of the end points of a diameter of a circle are given, we can also find the equation of the circle by finding the coordinates of the center and radius. The center is the mid-point of the diameter and radius is half of the length of the diameter.<\/p>\n
Examples :<\/h2>\n\n\n
Example : <\/span> Find the diameter form of the circle, the coordinates of the end points of whose diameter are (-1, 2) and (4, -3). <\/p>\n
Solution : <\/span>Given coordinates of two ends of diameter of circle i.e. (-1, 2) and (4, -3).
\nWe know that the equation of the circle described on the line segment joining \\((x_1, y_1)\\) and \\((x_2, y_2)\\) as a diameter is
\n(\\(x-x_1\\))(\\(x-x_2\\)) + (\\(y-y_1\\))(\\(y-y_2\\)) = 0
\nHere, \\(x_1\\) = -1, \\(x_2\\) = 4, \\(y_1\\) = 2 and \\(y_2\\) = -3
\nSo, the equation of the required circle is
\n(x + 1)(x – 4) + (y – 2)(y + 3) = 0
\n\\(x^2 + y^2 – 3x + y – 10\\) = 0
<\/p>\n\n\n\nExample : <\/span> Find the diameter form of the circle, drawn on the intercept made by the line 2x + 3y = 6 between the coordinates axes as diameter. <\/p>\n
Solution : <\/span>The line 2x + 3y = 6 meets X and Y axes at A(3, 0) and B(0, 2) respectively
\nWe know that the equation of the circle described on the line segment joining \\((x_1, y_1)\\) and \\((x_2, y_2)\\) as a diameter is
\n(\\(x-x_1\\))(\\(x-x_2\\)) + (\\(y-y_1\\))(\\(y-y_2\\)) = 0
\nTaking AB as a diameter, the equation of the required circle is
\n(x – 3)(x – 0) + (y – 0)(y – 2) = 0
\n\\(x^2 + y^2 – 3x – 2y\\) = 0
<\/p>\n\n\n\n