The area (in sq.

Question

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:

Accepted Answer

Equation of tangent to the
parabola y2 = 4x at P(1, 2),

T = 0

2y = $$4\left( {{{x + 1} \over 2}} \right)$$

$$ \Rightarrow $$ y = x + 1

Equation of normal of the tangent at point P(1, 2)

y - 2 = (-1)(x - 1)

$$ \Rightarrow $$ y - 2 = - x + 1

$$ \Rightarrow $$ x + y - 3 = 0

This normal also passes through the center (h, r) of the circle.

$$ \therefore $$ h + k - 3 = 0

$$ \Rightarrow $$ h = 3 - r

So center is (3 - r, r)

From picture you can see,

PC = r

$$ \Rightarrow $$ (PC)2 = r2

$$ \Rightarrow $$ (3 - r - 1)2 + (r - 2)2 = r2

$$ \Rightarrow $$ 4 + r2 - 4r + r2 + 4 - 4r = r2

$$ \Rightarrow $$ r2 - 8r + 8 = 0

$$ \therefore $$ r = $${{8 \pm \sqrt {64 - 32} } \over 2}$$

$$ \Rightarrow $$ r = $${{8 \pm \sqrt {32} } \over 2}$$

$$ \Rightarrow $$ r = $${{8 \pm 4\sqrt 2 } \over 2}$$

$$ \therefore $$ r = 4 + $${2\sqrt 2 }$$ and 4 - $${2\sqrt 2 }$$

If r = 4 + $${2\sqrt 2 }$$ then center of the circle is

(-1 - $${2\sqrt 2 }$$, 4 + $${2\sqrt 2 }$$). From the diagram you can see both the x coordinate and y coordinate of the circle should be positive but here x coordinate is negative.

So possible value of radius r = 4 - $${2\sqrt 2 }$$

Then area of the circle

= $$\pi $$r2

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

= $$\pi (16 + 8 - \[{16\sqrt 2 }$$)

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

Solutions · 1