question_answer

The area (in sq. units) of the smaller of the two circles that touch the parabola, \[{{y}^{2}}=4x\] at the point (1, 2) and the x-axis is :-             [JEE Main 9-4-2019 Afternoon]

A) \[4\pi (2-\sqrt{2})\]                 

B) \[8\pi (3-2\sqrt{2})\]

C) \[4\pi (3+\sqrt{2})\]               

D) \[8\pi (2-\sqrt{2})\]     

Correct Answer: B

Solution :

Equation of circle is \[{{(x-1)}^{2}}+{{(y-2)}^{2}}+\lambda (x-y+1)=0\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}+x(\lambda -2)+y(-4-\lambda )+(5+\lambda )=0\] As cirlce touches x axis then \[{{g}^{2}}-c=0\] \[\frac{{{(\lambda -2)}^{2}}}{4}=(5+\lambda )\] \[{{\lambda }^{2}}+4-4\lambda =20+4\lambda \] \[{{\lambda }^{2}}-8\lambda -16=0\] \[\lambda =\frac{8\pm \sqrt{128}}{2}\] \[\lambda =4\pm 4\sqrt{2}\] \[Radius=\left| \frac{(-4-\lambda )}{2} \right|\] Put \[\lambda \]and get least radius.