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Question

Let RS be the diameter of the circle x2+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).

A
(13,13)
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B
(14,12)
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C
(13,13)
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D
(14,12)
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Solution

The correct options are
A (13,13)
C (13,13)


Given equation of circle x2+y2=1
Let P(cosθ,sinθ) be any point on the circle
Tangent at P is
xcosθ+ysinθ1=0
Equation of line QS:x=1
Finding the point of intersection of line QS and the tangent, we get
cosθ+ysinθ1=0y=1cosθsinθQ=(1,1cosθsinθ)

Equation of normal at point P is
ycosθxsinθ+λ=0
Putting point P, we get
λ=0
Therefore, the equation of normal is
y=(tanθ) x
Line EQ:y=(1cosθsinθ)x

So, the point of intersection of normal and line EQ,
E=(1cosθsinθtanθ , 1cosθsinθ)E⎜ ⎜ ⎜tanθ2tanθ , tanθ2⎟ ⎟ ⎟

Assuming the point E=(h,k), we get
h=tanθ2tanθ, k=tanθ2
Now,
kh=tanθ=2tanθ21tan2θ2kh=2k1k2k3+2hkk=0k(k2+2h1)=0k2+2h1=0
Therefore, the locus of the point E is
2x+y21=0

By observation, we see that the points
(13,13) and (13,13) satisfies the locus.

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