935 views
The x-axis divides the circle x^2 + y^2 − 6x − 4y + 5 = 0 into two parts.
The area of the smaller part is

$x^2 + y^2 − 6x − 4y + 5 = 0$

$(x-3)^2+(y-2)^2+5-9-4=0$

$(x-3)^2+(y-2)^2=(\sqrt(8))^2$

It is a circle with center at $(3,2)$ and radius = $\sqrt8$.

The points of intersection of circle with x axis are $(1,0)$ and $(5,0)$. Rewrite the equation in terms of $\displaystyle\int y dx \\\displaystyle\sqrt{\left(8-(x-3)^2\right)}+2 = \int \sqrt{\left(8-(x-3)^2\right)} dx + \int2dx$ and so on..

Extending the comment by @Vineet Kumar 1  $OC$ is perpendicular on $AB$ and divides it in two equal parts

Let $\angle OAB = \theta$,

$cos\theta = \frac{OC}{OA} = \frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}$

$\theta = 45$

So, In $\triangle OAB,\angle OAB + \angle OBA + \angle AOB = 180$

Also, $\angle OAB = \angle OBA$. So, $\angle AOB=90$

So, Area of the required part $=$ Area of Sector $OAB$ $-$Area of triangle $OAB$

$= \pi r^2 * \frac{\angle AOB}{360}\,\,-\,\,\frac{1}{2}*OC*AB$

$= \pi (2\sqrt{2})^2 * \frac{90}{360} \,\,-\,\, \frac{1}{2}*2*4$

$=2.28$

Correct me if I'm wrong

correct

1 vote
1
3,799 views
1 vote