Question

The area of the smaller segment cut off from the circle ${{x}^{2}}+{{y}^{2}}=9$by $x=1$
A. $\dfrac{1}{2}\left( 9{{\sec }^{-1}}(3)-\sqrt{8} \right)$
B. $9{{\sec }^{-1}}(3)-\sqrt{8}$
C. $\sqrt{8}-9{{\sec }^{-1}}(3)$
D. None of there

Answer 1

Hint: In this question we need to find the area covered by the smallest region of the circle when cutting off the circle at point 1. To find this we need to find the central point of the given circle (at the origin) and also the length of the radius of the circle (from the center of the circle). As soon as both will get, we can find the smallest section of the circle out of two sections when cut off at point 1. As we get the section we will get limits (boundary points) to integrate the circle from a high limit to a small limit to get the area of the smallest section.

Formula Used: $\int {{x}^{n+1}}dx=\dfrac{{{x}^{n+1}}}{n+1}$

Complete step by step Solution:
Given the equation of the circle is,
${{x}^{2}}+{{y}^{2}}=9$
And this circle cut by the line $x=1$
The equation of the general circle with center at point h, k of X and Y coordinates and the radius r is given as
\[{{\left( x-h \right)}^{2}}+\text{ }{{\left( y-k \right)}^{2}}=\text{ }{{r}^{2}}\]
Where h, k is the X, Y coordinates of the center of given circle lies
By comparing the given equation with the standard or general equation of a circle we can conclude the value of h and k in this equation which is the central point of the center is 0 and 0. Thus, we can conclude the center of the circle laying at the origin of the three-axis such as
h = 0 and k = 0;
Also by comparing the given circle equation with the general equation we get the radius of the given circle is 3. Such as
${{(x+0)}^{2}}+{{(y+0)}^{2}}={{(3)}^{2}}$
${{x}^{2}}+{{y}^{2}}=9$
Now we need to find the smallest section when this given circle is cut off at 1, so that will the smallest section from point 1 to the total radius of the circle, 3 (not -3 otherwise it will be the largest section).

Now integrating the x-axis within the limit of 1 and 3 (upper limit is 3 and lower limit is 1) we get the area of the smallest section of the given circle from point 1 such as
  $y=\sqrt{9-{{x}^{2}}}$
\[\int\limits_{1}^{3}{\sqrt{9-{{x}^{2}}}}dx\]
\[\int\limits_{1}^{3}{x\sqrt{9-{{x}^{2}}}}+9{{\sin }^{-1}}\frac{x}{3}\]
\[\left[ x\sqrt{9-{{x}^{2}}}+9{{\sin }^{-1}}\frac{x}{3} \right]_{1}^{3}\]
\[\left[ 9\frac{\pi }{2}-\sqrt{8}-9{{\sin }^{-1}}\frac{1}{3} \right]\] (putting upper limit – lower limit)
\[\left[ 9\left[ \frac{\pi }{2}-{{\sin }^{-1}}\frac{1}{3} \right]-\sqrt{8} \right]\]
\[\left[ 9{{\cos }^{-1}}\frac{1}{3}-\sqrt{8} \right]\]

So, the integrated smallest region or total area of smallest region from \[x=1\]to \[x=3\]is \[\left[ 9{{\cos }^{-1}}\frac{1}{3}-\sqrt{8} \right]\]
So, $9{{\sec }^{-1}}(3)-\sqrt{8}$
Therefore, the correct option is (B).

Note: It is important to note that in the given question the point which is dividing the circle into two points is present on the x-axis due to which we integrate the circle within the limit of x and we find an equation to integrate into the term of x only like $y=\sqrt{9-{{x}^{2}}}$.