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  • Find the angle of intersection of the curves <img alt="x^{2}+y^{2}= 4\, and\, \left ( x-2 \right )^{2}+y^{2}= 4," src="https://entrancecorner.codecogs.com/gif.latex?x%5E%7B2%7D&plus;y%5E%7B2%7D%3D%204%5C%2C%20and%5C%2C%20%5Cleft%20%28%20x-2%20%5Cri

Find the angle of intersection of the curves x^{2}+y^{2}= 4\, and\, \left ( x-2 \right )^{2}+y^{2}= 4, at the point in the first quadrant.

 

 

 

 
 
 
 
 

Answers (1)

we have x^{2}+y^{2}= 4---(i)  & \left ( x-2 \right )^{2}+y^{2}= 4---(ii)
On solving (i) and (ii), we get : x^{2}-4x+4+y^{2}= 4
\Rightarrow x= 1\therefore y= \pm \sqrt{3}. so the points of intersection are:
P\left ( 1,\sqrt{3} \right )\, and\, Q\left ( 1,-\sqrt{3} \right )\cdotAs we have to find the angle of intersection in the firdt quadrant so, we shall take only point P.
Now for (i), 2x+2y\frac{dy}{dx}= 0\Rightarrow \frac{dy}{dx}= \frac{-x}{y}
\therefore slope at P= \frac{-1}{\sqrt{3}}
And for (ii) 2\left ( x-2 \right )+2y\frac{dy}{dx}= 0\Rightarrow \frac{dy}{dx}= \frac{-(x-2)}{y}
 \therefore slope at P= \frac{1}{\sqrt{3}}
Now the angle of intersection at point P
\tan \theta = \left | \frac{\frac{1}{\sqrt{3}}-\left ( \frac{-1}{\sqrt{3}} \right )}{1+\frac{1}{\sqrt{3}}\times \left ( \frac{-1}{\sqrt{3}} \right )} \right |= \left | \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} \right |= \sqrt{3}

Posted by

Ravindra Pindel

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