Question

Equation of the normal to the circle $x^{2}+y^{2}-2ax=0$ at the point $\left\{ a\left( 1+\cos \theta \right) ,a\sin \theta \right\} $ is given by
A) $y=\left( x-a\right) \tan \theta$
B) $y=\left( x+a\right) \cot \theta$
C) $y=x\tan \theta +a\cot \theta$
D) None of these

Answer 1

Hint: In this question it is given that we have to find the normal to the circle $x^{2}+y^{2}-2ax=0$ at the point $\left\{ a\left( 1+\cos \theta \right) ,a\sin \theta \right\} $. To find the solution we first need to find $\dfrac{dy}{dx}$ at the point $\left\{ a\left( 1+\cos \theta \right) ,a\sin \theta \right\} $. And as we know that the equation of normal at any point $\left( \alpha ,\beta \right) $ is
$$\left( x-\alpha \right) +\left[ \dfrac{dy}{dx} \right]_{\left( \alpha ,\beta \right) } \left( y-\beta \right) =0$$.............(1)

Complete step-by-step solution:
Given equation, $x^{2}+y^{2}-2ax=0$ ……………(2)
Differentiating both side of the above equation w.r.t ‘x’, we get,
$$\dfrac{d}{dx} \left( x^{2}+y^{2}-2ax\right) =0$$
$$\Rightarrow \dfrac{d}{dx} \left( x^{2}\right) +\dfrac{d}{dx} \left( y^{2}\right) -\dfrac{d}{dx} \left( 2ax\right) =0$$
$$\Rightarrow 2x+2y\dfrac{dy}{dx} -2a=0$$
$$\Rightarrow 2\left( x+y\dfrac{dy}{dx} -a\right) =0$$
$$\Rightarrow x+y\dfrac{dy}{dx} -a=0$$
$$\Rightarrow \dfrac{dy}{dx} =\dfrac{a-x}{y}$$
Now at point $\left\{ a\left( 1+\cos \theta \right) ,a\sin \theta \right\} $,
$$\left[ \dfrac{dy}{dx} \right]_{\left( a\left( 1+\cos \theta \right) ,a\sin \theta \right) } =\dfrac{a-a\left( 1+\cos \theta \right) }{a\sin \theta }$$
= $$\dfrac{a\left( 1-\left( 1+\cos \theta \right) \right) }{a\sin \theta }$$
= $$\dfrac{1-\left( 1+\cos \theta \right) }{\sin \theta }$$
= $$\dfrac{1-1-\cos \theta }{\sin \theta }$$
= $$\dfrac{-\cos \theta }{\sin \theta }$$.................(3)
Now by equation (1) we can write the equation of normal at $\left\{ a\left( 1+\cos \theta \right) ,a\sin \theta \right\} $ is,
$$\left\{ x-a\left( 1+\cos \theta \right) \right\} +\left( \dfrac{-\cos \theta }{\sin \theta } \right) \left( y-a\sin \theta \right) =0$$
$$\Rightarrow \left( x-a-a\cos \theta \right) -\left( \dfrac{\cos \theta }{\sin \theta } \right) \left( y-a\sin \theta \right) =0$$
$$\Rightarrow \left( x-a-a\cos \theta \right) =\left( \dfrac{\cos \theta }{\sin \theta } \right) \left( y-a\sin \theta \right) $$
Multiplying both side by $$\left( \dfrac{\sin \theta }{\cos \theta } \right) $$, we get,
$$\Rightarrow \left( x-a-a\cos \theta \right) \left( \dfrac{\sin \theta }{\cos \theta } \right) =\left( y-a\sin \theta \right) $$
$$\Rightarrow x\left( \dfrac{\sin \theta }{\cos \theta } \right) -a\left( \dfrac{\sin \theta }{\cos \theta } \right) -a\cos \theta \left( \dfrac{\sin \theta }{\cos \theta } \right) =\left( y-a\sin \theta \right) $$
$$\Rightarrow x\tan \theta -a\tan \theta -a\sin \theta =y-a\sin \theta$$
$$\left[ \because \dfrac{\sin \theta }{\cos \theta } =\tan \theta \right]$$
$$\Rightarrow x\tan \theta -a\tan \theta =y$$ [ by canceling $a\sin \theta$]
$$\Rightarrow y=\left( x-a\right) \tan \theta$$
Which is our required solution.
Thus the correct option is option A.

Note: To solve this type of question you need to memorise the equation of a normal line in any point on a curve which we have discussed earlier in the Hint portion. Also in any particular point on a circle we can draw only one normal line which always passes through the centre of the circle.