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Question: Equation of the normal to the circle \(x^{2}+y^{2}-2ax=0\) at the point \(\left\\{ a\left( 1+\cos \t...

Equation of the normal to the circle x2+y2−2ax=0x^{2}+y^{2}-2ax=0 at the point \left\\{ a\left( 1+\cos \theta \right) ,a\sin \theta \right\\} is given by
A) y=(x−a)tan⁡θy=\left( x-a\right) \tan \theta
B) y=(x+a)cot⁡θy=\left( x+a\right) \cot \theta
C) y=xtan⁡θ+acot⁡θy=x\tan \theta +a\cot \theta
D) None of these

Explanation

Solution

Hint: In this question it is given that we have to find the normal to the circle x2+y2−2ax=0x^{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 dydx\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 (α,β)\left( \alpha ,\beta \right) is
(x−α)+[dydx](α,β)(y−β)=0\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, x2+y2−2ax=0x^{2}+y^{2}-2ax=0 ……………(2)
Differentiating both side of the above equation w.r.t ‘x’, we get,
ddx(x2+y2−2ax)=0\dfrac{d}{dx} \left( x^{2}+y^{2}-2ax\right) =0
⇒ddx(x2)+ddx(y2)−ddx(2ax)=0\Rightarrow \dfrac{d}{dx} \left( x^{2}\right) +\dfrac{d}{dx} \left( y^{2}\right) -\dfrac{d}{dx} \left( 2ax\right) =0
⇒2x+2ydydx−2a=0\Rightarrow 2x+2y\dfrac{dy}{dx} -2a=0
⇒2(x+ydydx−a)=0\Rightarrow 2\left( x+y\dfrac{dy}{dx} -a\right) =0
⇒x+ydydx−a=0\Rightarrow x+y\dfrac{dy}{dx} -a=0
⇒dydx=a−xy\Rightarrow \dfrac{dy}{dx} =\dfrac{a-x}{y}
Now at point \left\\{ a\left( 1+\cos \theta \right) ,a\sin \theta \right\\} ,
[dydx](a(1+cos⁡θ),asin⁡θ)=a−a(1+cos⁡θ)asin⁡θ\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 }
= a(1−(1+cos⁡θ))asin⁡θ\dfrac{a\left( 1-\left( 1+\cos \theta \right) \right) }{a\sin \theta }
= 1−(1+cos⁡θ)sin⁡θ\dfrac{1-\left( 1+\cos \theta \right) }{\sin \theta }
= 1−1−cos⁡θsin⁡θ\dfrac{1-1-\cos \theta }{\sin \theta }
= −cos⁡θsin⁡θ\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
⇒(x−a−acos⁡θ)−(cos⁡θsin⁡θ)(y−asin⁡θ)=0\Rightarrow \left( x-a-a\cos \theta \right) -\left( \dfrac{\cos \theta }{\sin \theta } \right) \left( y-a\sin \theta \right) =0
⇒(x−a−acos⁡θ)=(cos⁡θsin⁡θ)(y−asin⁡θ)\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 (sin⁡θcos⁡θ)\left( \dfrac{\sin \theta }{\cos \theta } \right) , we get,
⇒(x−a−acos⁡θ)(sin⁡θcos⁡θ)=(y−asin⁡θ)\Rightarrow \left( x-a-a\cos \theta \right) \left( \dfrac{\sin \theta }{\cos \theta } \right) =\left( y-a\sin \theta \right)
⇒x(sin⁡θcos⁡θ)−a(sin⁡θcos⁡θ)−acos⁡θ(sin⁡θcos⁡θ)=(y−asin⁡θ)\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)
⇒xtan⁡θ−atan⁡θ−asin⁡θ=y−asin⁡θ\Rightarrow x\tan \theta -a\tan \theta -a\sin \theta =y-a\sin \theta
[∵sin⁡θcos⁡θ=tan⁡θ]\left[ \because \dfrac{\sin \theta }{\cos \theta } =\tan \theta \right]
⇒xtan⁡θ−atan⁡θ=y\Rightarrow x\tan \theta -a\tan \theta =y [ by canceling asin⁡θa\sin \theta]
⇒y=(x−a)tan⁡θ\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.