JEE Main & Advanced Mathematics Circle and System of Circles Normal to a Circle at a Given Point

  The normal of a circle at any point is a straight line, which is perpendicular to the tangent at the point and always passes through the centre of the circle.

(1) Equation of normal: The equation of normal to the circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] at any point \[({{x}_{1}},\,{{y}_{1}})\] is \[y-{{y}_{1}}=\frac{{{y}_{1}}+f}{{{x}_{1}}+g}(x-{{x}_{1}})\] or \[\frac{x-{{x}_{1}}}{{{x}_{1}}+g}=\frac{y-{{y}_{1}}}{{{y}_{1}}+f}\].

The equation of normal to the circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] at any point \[a{{x}^{2}}+2hxy+b{{y}^{2}}\] is \[x{{y}_{1}}-{{x}_{1}}y=0\] or \[\frac{x}{{{x}_{1}}}=\frac{y}{{{y}_{1}}}\].

(2) Parametric form : Since parametric co-ordinates of a point on the circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] is \[+2gx+2fy+c=0\].

\[\therefore \] Equation of normal at\[(a\,\,\cos \theta ,\,\,\,a\,\,\sin \theta )\] is \[\frac{x}{a\,\cos \theta }=\frac{y}{a\,\sin \,\theta }\] 

or  \[\frac{x}{\cos \theta }=\frac{y}{\sin \,\theta }\] or  \[y=x\,\,\tan \,\theta \]  or  \[y=mx\] where \[m=\tan \,\,\theta \], which is slope form of normal.