how to calculate the partial area of a circle given its radius?

Given: radius $r$ and two axis coordinates. Find areas $A$, $B$, $C$ and $D$

enter image description here


closed as off-topic by Saad, Namaste, ccorn, Xander Henderson, Dave Jun 13 '18 at 19:06

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    $\begingroup$ You can use integration. $\endgroup$ – John Glenn Jun 9 '18 at 3:32
  • $\begingroup$ Well you will need to declare another point $(x_c,y_c)$ to be the center of your circle, transform the equation of the circle with it's centre at the origin to the equation for that circle but with the origin at the point you have defined as $(x,y)$, rearrange the equation to the form $y=f(x)$ and then intergrate. The intervals of intergration for the four regions can be ascertained by elementary trigonometric and or vector analysis considerations regarding the distance between the centre of the circle and the point $(x,y)$, and the circle radius $r$. $\endgroup$ – Adam Jun 9 '18 at 4:03
  • $\begingroup$ This seems like homework,and so you can to the actual math yourself $\endgroup$ – Adam Jun 9 '18 at 4:05
  • $\begingroup$ user548331: not school homework, just for a script I am writing... Very cheeky you. $\endgroup$ – Erich Horn Jun 9 '18 at 4:09

You can always use integration. But if you want, you can always use geometry and algebra

You can always break the parts down into sectors and triangles. In the first quadrant, the area can be expressed as: $$A_{Q1}=A_{\text{sector}ADC}+A_{\triangle DIH}-A_{\triangle HAC}$$

enter image description here

The areas can be easily worked out if you just have the center $(h,k)$ and $r$ in the equation for your circle: $(x-h)^2+(y-k)^2=r^2$.

You can repeat the process for the rest of the quadrants, adding and subtracting areas as necessary.

For $Q2$: enter image description here

For $Q3$: enter image description here

For $Q4$: enter image description here


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