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QJEEM23Ma8300: JEE Main 2023 Mathematics - Class 11 JEEM23Ma8300 Solution: The points of intersection of the line a x + b y = 0 , ( a ≠ b…

The points of intersection of the line ax+by=0,(a≠b) and the circle x2+y2−2x=0 are A(α,0) and B(1,β). The image of the circle with AB as a diameter in the line x+y+2=0 is :

Multiple Choice Options

  • A x2+y2+5x+5y+12=0
  • B x2+y2+3x+5y+8=0
  • C x2+y2−5x−5y+12=0
  • D x2+y2+3x+3y+4=0

Step-by-Step Verified Solution

JEE Main 2023 (Online) 25th January Morning Shift Mathematics - Circle Question 38 English Explanation
As AA and BB satisfy both line and circle we have α=0⇒A(0,0)\alpha=0 \Rightarrow A(0,0) and β=1\beta=1 i.e. B(1,1)B(1,1)

Centre of circle as ABA B diameter is (12,12)\left(\frac{1}{2}, \frac{1}{2}\right) and radius =12=\frac{1}{\sqrt{2}}

∴\therefore For image of (12;12)\left(\frac{1}{2} ; \frac{1}{2}\right) in x+y+zx+y+z we get x−121=y−121=−2(3)2\frac{x-\frac{1}{2}}{1}=\frac{y-\frac{1}{2}}{1}=\frac{-2(3)}{2}

⇒\Rightarrow Image (−52,−52)\left(-\frac{5}{2},-\frac{5}{2}\right)

∴\therefore Equation of required circle

(x+52)2+(y+52)2=12\left(x+\frac{5}{2}\right)^{2}+\left(y+\frac{5}{2}\right)^{2}=\frac{1}{2}

⇒x2+y2+5x+5y+504−12=0\Rightarrow x^{2}+y^{2}+5 x+5 y+\frac{50}{4}-\frac{1}{2}=0

⇒x2+y2+5x+5y+12=0\Rightarrow x^{2}+y^{2}+5 x+5 y+12=0

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