Equation of the Normal to the Circle $x^2+y^2=a^2$ at the Point $(x_1,y_1)$ on the Circle
Let $P(x_1,y_1)$ be a point on the circle \[x^2+y^2=a^2\] Differentiating with respect to $x$, \[2x+2y\frac{dy}{dx}=0\] \[y\frac{dy}{dx}=-x\] \[\therefore\frac{dy}{dx}=-\frac{x}{y}\] Since the point $P(x_1,y_1)$ lies on the circle, \[\text{Slope of tangent }=\left|\frac{dy}{dx}\right|_{(x_1,y_1)}=-\frac{x_1}{y_1}\] \[\therefore\text{Slope of Normal }(m)=\frac{y_1}{x_1}\] Hence, the equation of the tangent line is, \[y-y_1=m(x-x_1)\] \[y-y_1=\frac{y_1}{x_1}(x-x_1)\] \[x_1y-x_1y_1=xy_1-x_1y_1\] \[x_1y=xy_1\] \[\therefore\frac{x}{x_1}=\frac{y}{y_1}\] This equation of the normal to the circle is in point form.
Normal to the General Circle $x^2+y^2+2gx+2fy+c=0$ at a Given Point on the Circle
Let $P(x_1,y_1)$ be a point on the circle \[x^2+y^2+2gx+2fy+c=0\]
Differentiating with respect to $x$, \[2x+2y\frac{dy}{dx}+2g+2f\frac{dy}{dx}+c=0\] \[(y+f)\frac{dy}{dx}=-(x+g)\] \[\frac{dy}{dx}=-\frac{x+g}{y+f}\]
Since the point $P(x_1,y_1)$ lies on the circle, \[\text{Slope of tangent}=\left|\frac{dy}{dx}\right|_{(x_1,y_1)}=-\frac{x_1+g}{y_1+f}\] \[\therefore\text{Slope of Normal }(m)=\frac{y_1+f}{x_1+g}\]
Hence the equation of the normal is, \[y-y_1=m(x-x_1)\] \[y-y_1=\frac{y_1+f}{x_1+g}(x-x_1)\] \[(y-y_1)(x_1+g)=(y_1+f)(x-x_1)\] \[x_1y+gy-x_1y_1-gy_1=xy_1-x_1y_1+fx-fx_1\] \[xy_1+fx-fx_1-x_1y-gy+gy_1=0\] \[(f+y_1)x-(g+x_1)y-fx_1+gy_1=0\] This is the equation of the normal in point form.
Equation of Normal to a Circle in Slope Form
Let $y=mx+c$ be a straight line and $x^2+y^2=a^2$ be a circle. We know that the equation of tangent to the circle in slope form is, \[y=m’x\pm a\sqrt{1+m’^2}\text{ }[m’=\text{Slope of Tangent}]\] [From: Condition of Tangency of a Straight Line to a Circle (Tangent to a Circle)]
Since normal is perpendicular to the tangent line, we have \[\text{Slope of normal }(m)=-\frac{1}{m’}\] \[\text{or, }m’=-\frac{1}{m}\] Hence the equation of normal is \[y=-\frac{1}{m}\pm a\sqrt{1+\frac{1}{m^2}}\] \[y=-\frac{1}{m}\pm a\sqrt{\frac{m^2+1}{m^2}}\] \[y=-\frac{1}{m}\pm\frac{a}{m}\sqrt{1+m^2}\] This is the required equation of normal in slope form.
Find the equation of normal to the circle $x^2+y^2=36$ at $(-6,0)$.
Given point $P(x_1,y_1)=(-6,0)$ and given circle is, \[x^2+y^2=36\] Now, the equation of normal line is, \[\frac{x}{x_1}=\frac{y}{y_1}\] \[\frac{x}{-6}=\frac{y}{0}\] \[\therefore y=0\] Hence, x-axis is normal to the given circle. [Know more: Equation of Straight Line Parallel to the Axes (Some Fundamental Formulae)]
Find the equation of normal to the circle $x^2+y^2-3x+10y-15=0$ at $(4,-11)$.
Given point $P(x_1,y_1)=(4,-11)$ and given circle is, \[x^2+y^2-3x+10y-15=0\] \[\begin{array}{c}\therefore 2g=-3, & 2f=10 &\text{and}& c=-15 \\ g=-\frac{3}{2}, & f=5 &\text{and}& c=-15\end{array}\] Hence, the equation of normal is, \[(f+y_1)x-(g+x_1)y-fx_1+gy_1=0\] \[(5-11)x-\left(-\frac{3}{2}+4\right)y-5×4+\frac{3}{2}×11=0\] \[-6x-\frac{5}{2}-20+\frac{33}{2}=0\] \[-12x-5y-40+33=0\] \[-12x-5y-7=0\] \[\therefore 12x+5y+7=0\]
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