Question

Minimum area of the circle which touches the parabolas $y=x^2+1$ and $y^2=x-1$ is

• A. $\frac{9\pi }{16}$ sq. units
• B. $\frac{9\pi }{32}$ sq. units
• C. $\frac{9\pi }{8}$ sq. units
• D. $\frac{9\pi }{4}$ sq. units

Answer 1

Answer: (B)

$\frac{9\pi }{32}$ sq. units

Parabola $y=x^2+1$
$x^2=y-1$
Parabola $x=y^2+1$
$y^2=x-1$
These parabolas are symmetrical about $y=x$.
Therefore, tangent at points of touch of parabola and circle are parallel to $y=x$.
Slope of tangent $=$ Slope of $y=x^2+1$ at point of touch
$\Rightarrow \frac{dy}{dx}=1$
$\Rightarrow 2x=1$
$\Rightarrow x=\frac{1}{2}$ and $y=\frac{5}{4}$
It's image about $y=x$ will be on $x=y^2+1$. Let the points be $A$ and $B$ respectively.
$A(\frac{1}{2},\frac{5}{4})$ and $B(\frac{5}{4},\frac{1}{2})$
$AB=\sqrt{{(\frac{1}{2}-\frac{5}{4})}^2+{(\frac{5}{4}-\frac{1}{2})}^2}=\sqrt{\frac{9}{16}+\frac{9}{16}}=\frac{3\sqrt{2}}{4}$
Radius $=\frac{1}{2}AB$
$=\frac{3}{8}\sqrt{2}$
Hence, area $=\pi (\frac{3}{8}\sqrt{2}{)}^2=\frac{9\pi }{32}$ sq.units