Q

# How to solve this problem- - Co-ordinate geometry - JEE Main-9

The area  ( in sq. units ) of the smaller of the two circle that touch the parabola , $y^{2}=4x$ at the point (1,2 ) and the x-axis is :

• Option 1)

$8\pi(2-\sqrt{2})$

• Option 2)

$4\pi (2-\sqrt{2})$

• Option 3)

$4\pi (3+\sqrt{2})$

• Option 4)

$8\pi (3-2\sqrt{2})$

Views

equation of tangent at (1,2)

$=2y=2(x+1)$

$y=x+1$

equation of normal

$y=-x+3$

let the centre be  $(3-r,r)$

$\\AC^{2}=r^{2}\\\\\:(3-r-1)^{2}+(r-2)^{2}=r^2$

$\\=(2-r)^{2}+(r-2)^{2}=r^{2}\\\\\:2(r-2)^{2}=r^{2}$

$\\2r^{2}-8r+8=r^{2}\\\\\:r^{2}-8r+8=0$

$r=4\pm 2\sqrt{2}$

for $r=4+ 2\sqrt{2}$

$3-r \:\:<0$

so

$r= 4-2\sqrt{2}$

$Area= \pir^{2}=8\pi(3-2\sqrt{2})$

Option 1)

$8\pi(2-\sqrt{2})$

Option 2)

$4\pi (2-\sqrt{2})$

Option 3)

$4\pi (3+\sqrt{2})$

Option 4)

$8\pi (3-2\sqrt{2})$

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