# 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.