Equation of Normal to a Circle with Examples

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.

Equation of Normal to a Circle :

(a) Equations of normal at point (\(x_1, y_1\)) of circle \({x}^2 + {y}^2 + 2gx + 2fy + c\) = 0 is

y – \(y_1\) = (\(y_1 + f\over {x_1 + g}\))(\(x – x_1\))

(b) The equations of normal on any point (\(x_1, y_1\)) of circle \(x^2\) + \(y^2\) = \(a^2\) is

\(y\over x\) = \(y_1\over x_1\)

(c) If \(x^2\) + \(y^2\) = \(a^2\) is the equation of the circle then at any point ‘t’ of this circle (acost, asint), the equation of normal is 

xsint – ycost = 0

Example : Find the normal to the circle \(x^2 + y^2\) = 0 at the point (1, 2).

Solution : We have, \(x^2 + y^2\) = 0

Since normal on any point (\(x_1, y_1\)) of circle \(x^2\) + \(y^2\) = \(a^2\) is \(y\over x\) = \(y_1\over x_1\).

i.e. \(y\over x\) = \(2\over 1\)

\(\implies\) y = 2x

which is the required normal to circle.

Example : Find the normal to the circle \(x^2 + y^2 – 5x + 2y – 48\) = 0 at the point (5, 6).

Solution : We have, \(x^2 + y^2 – 5x + 2y – 48\) = 0

Since 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)

i.e. y + 1 = \(7\over {5/2}\)(x – \(5\over 2\)) \(\implies\) 5y + 5 = 14x – 35

\(\implies\) 14x – 5y – 40 = 0

which is the required normal to circle.

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