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JEE Main 202126 AugustEvening Shift
Exam
JEE Main
Year
2021
Date
26 August
Shift
Evening Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2021 · 26 August · Evening Shift · Mathematics · Circle

A circle C touches the line x = 2y at the point (2, 1) and...

A circle CC touches the line x=2yx = 2y at the point (2,1)\left(2, 1\right) and intersects the circle C1:x2+y2+2y−5=0{C}_{1}: {x}^{2} + {y}^{2} + 2y - 5 = 0 at two points PandQP and Q such that PQPQ is a diameter of C1{C}_{1}. Then the diameter of CC is :

Options

  1. A4154\sqrt{15}
  2. B285\sqrt{285}
  3. C1515
  4. D757\sqrt{5}

Correct answer

D. 757\sqrt{5}

Step-by-step solution

Family of circle touching line 2y=x2y = x at point \left(2, 1\right)<br/><br />( x - 2{)}^{2} + ( y - 1{)}^{2} + \lambda ( x - 2y ) = 0 . . . ( 1 )<br/>commonchord<br />common chordPQis<br/>is<br />( x - 2{)}^{2} + ( y - 1{)}^{2} + \lambda ( x - 2y ) - {x}^{2} - {y}^{2} - 2y + 5 = 0<br/>(diameterof<br />(diameter of{C}_{1}passesthroughpasses through( 0, - 1 ))<br/>)<br />4 + 4 + 2\lambda - 1 + 2 + 5 = 0</p><p></p><p>\lambda = - 7putinequation(1)put in equation \left(1\right)

(x−2)2+(y−1)2−7(x−2y)=0( x - 2{)}^{2} + ( y - 1{)}^{2} - 7 ( x - 2y ) = 0

x2+y2−11x+12y+5=0{x}^{2} + {y}^{2} - 11x + 12y + 5 = 0

r=1214+36−5=121+1244=2454=2452⇒diameter=245=75r = \sqrt{\dfrac{121}{4} + 36 - 5} = \sqrt{\dfrac{121 + 124}{4}} = \sqrt{\dfrac{245}{4}} = \dfrac{\sqrt{245}}{2} ⇒ \text{diameter} = \sqrt{245}= 7\sqrt{5}

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