{"id":3434,"date":"2021-07-24T13:14:43","date_gmt":"2021-07-24T13:14:43","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3434"},"modified":"2021-11-20T22:39:29","modified_gmt":"2021-11-20T17:09:29","slug":"equation-of-normal-to-circle","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-normal-to-circle\/","title":{"rendered":"Equation of Normal to a Circle with Examples"},"content":{"rendered":"
The normal at a point is the straight line which is perpendicular to the tangent to circle at the point of contact. Normal at a point of the circle passes through the center of circle. Here, you will learn how to find equation of normal to a circle with example.<\/p>\n
(a) Equations of normal at point (\\(x_1, y_1\\)) of circle \\({x}^2 + {y}^2 + 2gx + 2fy + c\\) = 0 is<\/p>\n
\ny – \\(y_1\\) = (\\(y_1 + f\\over {x_1 + g}\\))(\\(x – x_1\\))<\/strong><\/p>\n<\/blockquote>\n
(b) The equations of normal on any point (\\(x_1, y_1\\)) of circle \\(x^2\\) + \\(y^2\\) = \\(a^2\\) is<\/p>\n
\n\\(y\\over x\\) = \\(y_1\\over x_1\\)<\/strong><\/p>\n<\/blockquote>\n
(c) If \\(x^2\\) + \\(y^2\\) = \\(a^2\\) is the equation of the circle then at any point ‘t’ of this circle (acost, asint)<\/strong>, the equation of normal is <\/p>\n
\nxsint – ycost = 0<\/strong><\/p>\n<\/blockquote>\n\n\n
Example : <\/span> Find the normal to the circle \\(x^2 + y^2\\) = 0 at the point (1, 2).<\/p>\n
Solution : <\/span>We have, \\(x^2 + y^2\\) = 0
\nSince normal on any point (\\(x_1, y_1\\)) of circle \\(x^2\\) + \\(y^2\\) = \\(a^2\\) is \\(y\\over x\\) = \\(y_1\\over x_1\\).
\ni.e. \\(y\\over x\\) = \\(2\\over 1\\)
\n\\(\\implies\\) y = 2x
\nwhich is the required normal to circle.
<\/p>\n\n\n\nExample : <\/span> Find the normal to the circle \\(x^2 + y^2 – 5x + 2y – 48\\) = 0 at the point (5, 6).<\/p>\n
Solution : <\/span>We have, \\(x^2 + y^2 – 5x + 2y – 48\\) = 0
\nSince the normal to the circle always passes through center so equation of the normal will be the line passing through (5,6) & (\\(5\\over 2\\), -1)
\ni.e. y + 1 = \\(7\\over {5\/2}\\)(x – \\(5\\over 2\\)) \\(\\implies\\) 5y + 5 = 14x – 35
\n\\(\\implies\\) 14x – 5y – 40 = 0
\nwhich is the required normal to circle.
<\/p>\n\n\n\n