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Circle question

2016 · Shift 1 · Q28
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  5. /2016 · Shift 1 · Q28

Circle question

2016 · Shift 1 · Q28

JEE AdvancedMathematicsCircleMultiple correct+4 / −2
Let RS be the diameter of the circle x2 + y2=1{x^2}\, + \,{y^2} = 1x2+y2=1, where S is the point (1, 0). Let P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. Then the locus of E passes through the point (s)
  1. A
    (13 ,13)\left( {{1 \over 3}\,,{1 \over {\sqrt 3 }}} \right)(31​,3​1​)
  2. B
    (14 ,12)\left( {{1 \over 4}\,,{1 \over 2}} \right)(41​,21​)
  3. C
    (13 ,−13)\left( {{1 \over 3}\,, - {1 \over {\sqrt 3 }}} \right)(31​,−3​1​)
  4. D
    (14 ,−12)\left( {{1 \over 4}\,,-{1 \over 2}} \right)(41​,−21​)
View written solutionFree

Correct answer: A, C

  1. Set up the circle and points

The circle is x2+y2=1x^2+y^2=1x2+y2=1 with center at the origin.

Since S=(1,0)S=(1,0)S=(1,0) and RSRSRS is a diameter, the opposite end is R=(−1,0).R=(-1,0).R=(−1,0). So the line RSRSRS is the xxx-axis.

Let the variable point PPP on the circle be P=(cos⁡θ,sin⁡θ),P=(\cos\theta,\sin\theta),P=(cosθ,sinθ), where P≠R,SP\neq R,SP=R,S, so sin⁡θ≠0\sin\theta\neq 0sinθ=0 and cos⁡θ≠±1\cos\theta\neq \pm 1cosθ=±1 as needed.

  1. Find the tangents at SSS and PPP

For the circle x2+y2=1x^2+y^2=1x2+y2=1, the tangent at (x1,y1)(x_1,y_1)(x1​,y1​) is xx1+yy1=1.xx_1+yy_1=1.xx1​+yy1​=1.

  • Tangent at S=(1,0)S=(1,0)S=(1,0): x=1.x=1. x=1.

  • Tangent at P=(cos⁡θ,sin⁡θ)P=(\cos\theta,\sin\theta)P=(cosθ,sinθ): xcos⁡θ+ysin⁡θ=1.x\cos\theta+y\sin\theta=1.xcosθ+ysinθ=1.

These two tangents meet at QQQ.

Since QQQ lies on x=1x=1x=1, substitute x=1x=1x=1 into the tangent at PPP: cos⁡θ+ysin⁡θ=1\cos\theta+y\sin\theta=1cosθ+ysinθ=1 ysin⁡θ=1−cos⁡θy\sin\theta=1-\cos\thetaysinθ=1−cosθ y=1−cos⁡θsin⁡θ=tan⁡θ2.y=\frac{1-\cos\theta}{\sin\theta}=\tan\frac\theta2.y=sinθ1−cosθ​=tan2θ​.

Hence, Q=(1,tan⁡θ2).Q=\left(1,\tan\frac\theta2\right).Q=(1,tan2θ​).

  1. Equation of the normal at PPP

For a circle, the normal at any point passes through the center. So the normal at PPP is the line through the origin and P=(cos⁡θ,sin⁡θ)P=(\cos\theta,\sin\theta)P=(cosθ,sinθ).

Thus its equation is y=(tan⁡θ)xy=(\tan\theta)xy=(tanθ)x provided cos⁡θ≠0\cos\theta\neq 0cosθ=0; equivalently, in parametric form, (x,y)=λ(cos⁡θ,sin⁡θ).(x,y)=\lambda(\cos\theta,\sin\theta).(x,y)=λ(cosθ,sinθ).

  1. Equation of the line through QQQ parallel to RSRSRS

Since RSRSRS is the xxx-axis, a line through QQQ parallel to RSRSRS is horizontal: y=tan⁡θ2.y=\tan\frac\theta2.y=tan2θ​.

This line meets the normal at PPP at EEE.

  1. Find coordinates of EEE

Point EEE lies on the normal, so let (x,y)=λ(cos⁡θ,sin⁡θ).(x,y)=\lambda(\cos\theta,\sin\theta).(x,y)=λ(cosθ,sinθ).

Also, since EEE lies on the horizontal line through QQQ, y=tan⁡θ2.y=\tan\frac\theta2.y=tan2θ​.

Thus, λsin⁡θ=tan⁡θ2.\lambda\sin\theta=\tan\frac\theta2.λsinθ=tan2θ​. Using tan⁡θ2=sin⁡θ1+cos⁡θ,\tan\frac\theta2=\frac{\sin\theta}{1+\cos\theta},tan2θ​=1+cosθsinθ​, we get λsin⁡θ=sin⁡θ1+cos⁡θ\lambda\sin\theta=\frac{\sin\theta}{1+\cos\theta}λsinθ=1+cosθsinθ​ so λ=11+cos⁡θ.\lambda=\frac1{1+\cos\theta}.λ=1+cosθ1​.

Therefore, x=λcos⁡θ=cos⁡θ1+cos⁡θ,x=\lambda\cos\theta=\frac{\cos\theta}{1+\cos\theta},x=λcosθ=1+cosθcosθ​, y=λsin⁡θ=sin⁡θ1+cos⁡θ=tan⁡θ2.y=\lambda\sin\theta=\frac{\sin\theta}{1+\cos\theta}=\tan\frac\theta2.y=λsinθ=1+cosθsinθ​=tan2θ​.

So the locus is parametrically E(cos⁡θ1+cos⁡θ,sin⁡θ1+cos⁡θ).E\left(\frac{\cos\theta}{1+\cos\theta},\frac{\sin\theta}{1+\cos\theta}\right).E(1+cosθcosθ​,1+cosθsinθ​).

  1. Eliminate the parameter

Let the coordinates of EEE be (x,y)(x,y)(x,y). Then x=cos⁡θ1+cos⁡θ,y=sin⁡θ1+cos⁡θ.x=\frac{\cos\theta}{1+\cos\theta}, \qquad y=\frac{\sin\theta}{1+\cos\theta}. x=1+cosθcosθ​,y=1+cosθsinθ​.

From the first equation, x(1+cos⁡θ)=cos⁡θx(1+\cos\theta)=\cos\thetax(1+cosθ)=cosθ x=cos⁡θ(1−x)x=\cos\theta(1-x)x=cosθ(1−x) cos⁡θ=x1−x.\cos\theta=\frac{x}{1-x}. cosθ=1−xx​.

From the second, y=sin⁡θ1+cos⁡θ.y=\frac{\sin\theta}{1+\cos\theta}. y=1+cosθsinθ​.

Now use sin⁡2θ+cos⁡2θ=1.\sin^2\theta+\cos^2\theta=1.sin2θ+cos2θ=1. A simpler way is to use the half-angle substitution: t=tan⁡θ2=y.t=\tan\frac\theta2=y.t=tan2θ​=y. Then

\qquad \sin\theta=\frac{2t}{1+t^2}.$$ So $$x=\frac{\frac{1-t^2}{1+t^2}}{1+\frac{1-t^2}{1+t^2}} =\frac{1-t^2}{2}.$$ Since $t=y$, $$x=\frac{1-y^2}{2}.$$ Hence the locus is $$y^2=1-2x.$$ --- 7. **Check which given points lie on the locus** We test each option in $$y^2=1-2x.$$ ### Option A: $\left(\frac13,\frac1{\sqrt3}\right)$ LHS: $$y^2=\left(\frac1{\sqrt3}\right)^2=\frac13.$$ RHS: $$1-2x=1-2\cdot\frac13=1-\frac23=\frac13.$$ So A lies on the locus. ### Option B: $\left(\frac14,\frac12\right)$ LHS: $$y^2=\frac14.$$ RHS: $$1-2x=1-2\cdot\frac14=\frac12.$$ Not equal, so B does not lie on the locus. ### Option C: $\left(\frac13,-\frac1{\sqrt3}\right)$ LHS: $$y^2=\left(-\frac1{\sqrt3}\right)^2=\frac13.$$ RHS: $$1-2x=1-\frac23=\frac13.$$ So C lies on the locus. ### Option D: $\left(\frac14,-\frac12\right)$ LHS: $$y^2=\frac14.$$ RHS: $$1-2x=\frac12.$$ Not equal, so D does not lie on the locus. --- 8. **Final answer** The locus passes through: $$\left(\frac13,\frac1{\sqrt3}\right), \quad \left(\frac13,-\frac1{\sqrt3}\right).$$ So the correct options are **A and C**.
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