Question

Tangent and Normal
12. If the straight line lx+my+n=0 touches the :
(i) circle $x^2+y^2=a^2,showthat,a^2(l^2+m^2)=n^2$

Answer 1

$$\begin{gathered} C:x^2+y^2=a^2 \\ \frac{d{x}^2+y^2}{dx}=\frac{d{a}^2}{dx} \\ \frac{d\left(x^2\right)}{dx}+\frac{d\left(y^2\right)}{dx}=0 \\ 2x+2y\times \frac{dy}{dx}=0 \\ \frac{dy}{dx}=-\frac{x}{y} \\ LetthepointoncirclebeP(x_1,y_1) \\ {\left.\frac{dy}{dx}\right\}}_{atP(x1,y1)}=-\frac{{\displaystyle x_1}}{{\displaystyle y_1}} \\ Eqationof\tan gentatP \\ (y-y_1)=-\frac{x_1}{y_1}(x-x_1) \\ y{y}_1-{y_1}^2={x_1}^2-x{x}_1 \\ y{y}_1+x{x}_1={x_1}^2+{y_1}^2 \\ \frac{{\displaystyle x_1}}{{x_1}^2+{y_1}^2}x+\frac{{\displaystyle y_1}}{{x_1}^2+{y_1}^2}y=1 \\ lx+my=n \\ -\frac{l}{n}x-\frac{m}{n}y=1 \\ Comparingtheaboveandderivedequation \\ \frac{{\displaystyle l}}{{\displaystyle n}}=-\frac{{\displaystyle x_1}}{{x_1}^2+{y_1}^2}\_\_\_\_\_\_\_\_\_\_\_\_\_\left(1\right) \\ \frac{{\displaystyle m}}{{\displaystyle n}}=-\frac{{\displaystyle y_1}}{{x_1}^2+{y_1}^2}\_\_\_\_\_\_\_\_\_\_\_\_\_\left(2\right) \\ Squaringandaddingequation1and2 \\ \frac{{\displaystyle l^2}}{{\displaystyle n^2}}+\frac{{\displaystyle m^2}}{{\displaystyle n^2}}=\frac{{\displaystyle {x_1}^2}}{({x_1}^2+{y_1}^2{)}^2}+\frac{{\displaystyle {y_1}^2}}{({x_1}^2+{y_1}^2{)}^2}=\frac{{\displaystyle 1}}{({x_1}^2+{y_1}^2)}\_\_\_\_\_\_\_\_\_\_\left(3\right) \\ PutP(x_1,y_1)inequationofcircleasitliesonit \\ {x_1}^2+{y_1}^2=a^2 \\ Putinequation3 \\ \frac{{\displaystyle l^2}}{{\displaystyle n^2}}+\frac{{\displaystyle m^2}}{{\displaystyle n^2}}=\frac{1}{a^2} \\ {a}^{\mathbf{2}}{\mathit{l}}^{\mathbf{2}}\mathbf{+}{\mathit{a}}^{\mathbf{2}}{\mathit{m}}^{\mathbf{2}}\mathbf{=}{\mathit{n}}^{\mathbf{2}} \end{gathered}$$