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QJEEAdv18Ma3536: JEE Advanced 2018 Mathematics - Class 11 JEEAdv18Ma3536 Solution: Let T be the line passing through the points P( 2, 7) and Q(2,…

Let T be the line passing through the points P( −- 2, 7) and Q(2, −- 5). Let F1 be the set of al pairs of circles (S1, S2) such that T is tangent to S1 at P and tangent to S2 at Q, and also such that S1 and S2 touch each other at a point, say M. Let E1 be the set representing the locus of M as the pair (S1, S2) varies in F1. Let the set of all straight line segments joining a pair of distinct points of E1 and passing through the point R(1, 1) be F2. Let E2 be the set of the mid-points of the line segments in the set F2. Then, which of the following statement(s) is (are) TRUE?

Multiple Choice Options

  • A The point ( −- 2, 7) lies in E1
  • B The point (45,75)\left( {{4 \over 5},{7 \over 5}} \right) does not lie in E2
  • C The point (12,1)\left( {{1 \over 2},1} \right) lies in E2
  • D The point (0,32)\left( {0,{3 \over 2}} \right) does not lie in E1

Step-by-Step Verified Solution

It is given that T is tangents to S1 at P and S2 at Q and S1 and S2 touch externally at M.

JEE Advanced 2018 Paper 2 Offline Mathematics - Hyperbola Question 6 English Explanation

∴\therefore MN = NP = NQ

∴\therefore Locus of M is a circle having PQ as its diameter of circle

∴\therefore Equation of circle

(x−2)(x+2)+(y+5)(y−7)=0(x - 2)(x + 2) + (y + 5)(y - 7) = 0

⇒x2+y2−2y−39=0\Rightarrow {x^2} + {y^2} - 2y - 39 = 0

Hence,

E1:x2+y2−2y−39=0{E_1}:{x^2} + {y^2} - 2y - 39 = 0 , x≠±2x \ne \pm 2

Locus of mid-point of chord (h, k) of the circle E1 is

xh+yk−(y+k)−39=h2+k2−2k−39xh + yk - (y + k) - 39 = {h^2} + {k^2} - 2k - 39

⇒xh+yk−y−k=h2+k2−2k\Rightarrow xh + yk - y - k = {h^2} + {k^2} - 2k

Since, chord is passing through (1, 1).

∴\therefore Locus of mid-point of chord (h, k) is

h+k−1−k=h2+k2−2kh + k - 1 - k = {h^2} + {k^2} - 2k

⇒h2+k2−2k−h+1=0\Rightarrow {h^2} + {k^2} - 2k - h + 1 = 0

Locus is E2:x2+y2−x−2y+1=0{E_2}:{x^2} + {y^2} - x - 2y + 1 = 0

Now, after checking options, (a) and (d) are correct.

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