Carol Gates

2021-09-25

Find the overlapping area of two equations also give its geometrical representation:
${u}^{2}+{v}^{2}=n{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{u}^{2}+{(v-n)}^{2}=1$
put $n=516$

Viktor Wiley

Skilled2021-09-26Added 84 answers

Step 1
The equation of the form $x}^{2}+{y}^{2}={r}^{2$ is the standard equation of a circle with radius r. To obtain the overlapping area the intersection of the circles should be obtained.
Step 2
The function ${u}^{2}+{v}^{2}=nf{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}n=516$ gives a circle with center at (0,0) and radius approximately 22.7156. The equation ${u}^{2}+{(v-n)}^{2}=1f{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}n=516$ gives an circle with center at (0, 516) and radius 1. Therefore, these circles do not intersect and hence the overlapping area is 0.
Step 3
These equations do not intersect and hence the overlapping area is 0.

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