Verified Solution • 2025–26 Session moderate

QJEEAdv19Ma6e26: JEE Advanced 2019 Mathematics - Class 11 JEEAdv19Ma6e26 Solution: A line y = mx + 1 intersects the circle = 25 at the points P…

A line y = mx + 1 intersects the circle (x−3)2+(y+2)2{(x - 3)^2} + {(y + 2)^2} = 25 at the points P and Q. If the midpoint of the line segment PQ has x-coordinate −35- {3 \over 5} , then which one of the following options is correct?

Multiple Choice Options

  • A 6 ≤\le m < 8
  • B −- 3 ≤\le m < −- 1
  • C 4 ≤\le m < 6
  • D 2 ≤\le m < 4

Step-by-Step Verified Solution

It is given that points P and Q are intersecting points of circle (x−3)2+(y+2)2{(x - 3)^2} + {(y + 2)^2} = 25 .....(i)

Line y = mx + 1 .....(ii)

And, the mid-point of PQ is A having x-coordinate −35- {3 \over 5}

so y-coordinate is 1−351 - {3 \over 5} m.

So, A(−35,1−35m)A\left( { - {3 \over 5},1 - {3 \over 5}m} \right)

From the figure,

∵\because AC ⊥\bot PQ

JEE Advanced 2019 Paper 1 Offline Mathematics - Circle Question 14 English Explanation

⇒\Rightarrow (slope of AC) ×\times (slope of PQ) = −- 1

⇒\Rightarrow (−2−1+35m3+35)×m=−1\left( {{{ - 2 - 1 + {3 \over 5}m} \over {3 + {3 \over 5}}}} \right) \times m = - 1

⇒(3/5)m−318/5m=−1\Rightarrow {{(3/5)m - 3} \over {18/5}}m = - 1

⇒(3m−1518)m=−1\Rightarrow \left( {{{3m - 15} \over {18}}} \right)m = - 1

⇒3m2−15m+18=0\Rightarrow 3{m^2} - 15m + 18 = 0

⇒m2−5m+6=0\Rightarrow {m^2} - 5m + 6 = 0

⇒\Rightarrow m = 2 or 3

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