How to find the intersection points inside circle

Question

I would like to find the intersection points inside a circle for compute the area.

The circle has a region R inside the circle $x^2 + y^2 = 4$ and above the line $y = x + 2$ by a definite integral related to $x$.

I did:

1) $x^2 + y^2 = 4$

2) $y^2 = 4 - x^2$

3) $y = \sqrt{4 - x^2}$

then

4) $x + 2 = \sqrt{4-x^2}$

5) $x + 2 = 2 - x$

6) $x + x = 0$

7) $2x = 0$

8) $x = 0$

How to calculate the intersection points? The answers is that the integral goes from -2 to 0. How to find it?

I always do that correctly. But this case is different.

Answer 1

We should find the intersection points between the line $y=x+2$ and the circle $x^2+y^2=4$. With the solutions you had, step 2 going step 3 is wrong. Instead, we proceed as follows: $$\begin{align}x^2+(x+2)^2=4&\iff x^2+x^2+4x+4=4\\ &\iff 2x^2+4x=0\\ &\iff x^2+2x=0\\ &\iff x(x+2)=0\\ &\iff x=0\text{ or }x=-2. \end{align}$$ If $x=0$ then $y=x+2=0+2=2$. If $x=-2$ then $y=x+2=-2+2=0$. Hence, the intersection points are $(-2,0)$ and $(0,2)$.

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Since you mention about an integral, maybe you want the following:

$$A=\int_{-2}^0\big[(\sqrt{4-x^2})-(x+2)\big]dx$$

Answer 2

In addition to the other answers that give the solution, I want to point out where you went awry:

• Going from item 2) to 3), you should have $y = \pm \sqrt{4-x^2}$. This is similar to $x^2 = 9 \iff x = \pm 3$, not just $x=3$.

• Going from item 4) to 5), $\sqrt{4- x^2} \neq 2 -x$, it is just $\sqrt{4-x^2}$. In general, $\sqrt{a^2 + b^2} \neq a + b$. Consider $$5 = \sqrt{25} = \sqrt{4^2 + 3^2} \neq \sqrt{4^2} + \sqrt{3^2} = 4+3 = 7.$$