How do we find equation of tangent and normal to a circle at a given point?

Apr 11, 2017

Equation of tangent at $\left({x}_{1} , {y}_{1}\right)$ at circle ${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$ is $x {x}_{1} + y {y}_{1} + g \left(x + {x}_{1}\right) + f \left(y + {y}_{1}\right) + c = 0$ and equation of normal is $\left({y}_{1} + f\right) x - \left({x}_{1} + g\right) y - f {x}_{1} + g {y}_{1} = 0$

Explanation:

Let the circle be ${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$,

let us seek normal and tangent at $\left({x}_{1} , {y}_{1}\right)$

(note that ${x}_{1}^{2} + {y}_{1}^{2} + 2 g {x}_{1} + 2 f {y}_{1} + c = 0$) ...(A)

as the center of the circle is $\left(- g , - f\right)$,

as normal joins $\left(- g , - f\right)$ and $\left({x}_{1} , {y}_{1}\right)$ its slope is $\frac{{y}_{1} + f}{{x}_{1} + g}$

and equation of normal is $y - {y}_{1} = \frac{{y}_{1} + f}{{x}_{1} + g} \left(x - {x}_{1}\right)$

i.e. $\left({x}_{1} + g\right) y - {x}_{1} {y}_{1} - g {y}_{1} = \left({y}_{1} + f\right) x - {x}_{1} {y}_{1} - f {x}_{1}$

or $\left({y}_{1} + f\right) x - \left({x}_{1} + g\right) y - f {x}_{1} + g {y}_{1} = 0$

or $\left({y}_{1} + f\right) x - \left({x}_{1} + g\right) y - f {x}_{1} + g {y}_{1} = 0$

and as product of slopes of normal and tangent is $- 1$,

slope of tangent is $- \frac{{x}_{1} + g}{{y}_{1} + f}$ and its equation will be

$y - {y}_{1} = - \frac{{x}_{1} + g}{{y}_{1} + f} \left(x - {x}_{1}\right)$

or $\left(y - {y}_{1}\right) \left({y}_{1} + f\right) + \left(x - {x}_{1}\right) \left({x}_{1} + g\right) = 0$

or $y {y}_{1} - {y}_{1}^{2} + f y - f {y}_{1} + x {x}_{1} - {x}_{1}^{2} + g x - g {x}_{1} = 0$

or $x {x}_{1} + y {y}_{1} + g x + f y - \left({x}_{1}^{2} + {y}_{1}^{2} + f {y}_{1} + g {x}_{1}\right) = 0$

from (A) ${x}_{1}^{2} + {y}_{1}^{2} + f {y}_{1} + g {x}_{1} = - f {y}_{1} - g {x}_{1} - c$

Hence equation of tangent is $x {x}_{1} + y {y}_{1} + g x + f y + f {y}_{1} + g {x}_{1} + c = 0$

or $x {x}_{1} + y {y}_{1} + g \left(x + {x}_{1}\right) + f \left(y + {y}_{1}\right) + c = 0$