Intersection of two circles

In a nutshell

To find the point at which two circles intersect, the equations for the two circles may be solved simultaneously. 

Example 1

The circle with equation $$(x-11)^2+(y-2)^2=100$$ and centre $$A$$ intersects the circle with equation $$(x-7)^2+(y-4)^2=40$$ and centre $$B$$ and points $$R$$ and $$S$$.

a) Find the intersection points $$R$$​ and $$S$$.

First, expand the equation for the circle with centre $$A$$, and rearrange to make the equation equal to $$0$$.

$$\begin{aligned}(x-11)^2 + (y-2)^2 &= 100\\x^2-22x+121+y^2-4y+4&=100\\x^2-22x+y^2-4y+25&=0 \end{aligned}$$

Do the same for circle with centre $$B$$.

$$\begin{aligned}(x-7)^2+(y-4)^2&=40 \\x^2-14x+49+y^2-8y+16&=40\\x^2-14x+y^2-8y+25&=0 \end{aligned}$$​​

Solve the equations and make $$y$$ the subject. ​

$$\begin{aligned}x^2-22x+y^2-4y+25&=x^2-14x+y^2-8y+25 \\-22x-4y+25&=-14x-8y+25\\4y&=8x \\y&=2x \end{aligned}$$​​

Now, substitute the above equation into $$x^2-22x+y^2-4y+25=0$$.

$$\begin{aligned}x^2-22x+(2x)^2-4(2x)+25&=0 \\x^2-22x+4x^2-8x+25&=0 \\5x^2-30x+2&5=0 \\x&=\dfrac{--30\pm\sqrt{(-30)^2-(4\times5\times25)}}{2\times5} \\\\x&= \dfrac{30\pm\sqrt{900-500}}{10} \\\\x&= \dfrac{30\pm\sqrt{400}}{10} \\\\x&=5, 1 \end{aligned}$$​​

Finally, substitute values for $$x$$​ into $$y=2x$$​ to solve for $$y$$ to find points $$R$$ and $$S$$.

$$\begin{aligned}y&=2(5)\\y&=10\\\\y&=2(1)\\y&=2\end{aligned}$$

Therefore, points $$R$$ and $$S$$ are $$\underline{(5,10)}$$ and $$\underline{(1,2)}$$, respectively.​

b) Find the area of the right-angled triangle $$RBS$$.

To find the area of the triangle, the lengths of the sides $$BR, BS$$ and $$RS$$ must be found. As points $$R$$ and $$S$$ lie on the circle with centre $$B$$.

$$BR=BS=r = \sqrt{40} = 2\sqrt{10}$$​​

To find the length of $$RS$$, use $$d=\sqrt{{(x_{2} - x_{1}})^2+{(y_{2} - y_{1})^2}}$$ for points $$R(5,10)$$ and $$S(1,2)$$.

  $$\begin{aligned}d&=\sqrt{{(x_{2} - x_{1}})^2+{(y_{2} - y_{1})^2}} \\&=\sqrt{{(1 - 5})^2+{(2 - 10)^2}}\\&=\sqrt{80} \\&=4\sqrt5\end{aligned}$$

Use the cosine rule to find the angle $$\angle RBS$$:

$$\begin{aligned}\cos x &= \dfrac {b^2+c^2-a^2}{2bc} \\\\\cos x&= \dfrac{(2\sqrt{10})^2 + (2\sqrt{10})^2 - (4\sqrt{5})^2}{2 \times2\sqrt{10} \times 2\sqrt{10} } \\\\ \cos x &=0 \\\\x &= \cos^{-1} (0) \\x&= 90 \degree\end{aligned}$$​​

As triangle $$RBS$$ is right angled, find the area using $$A = \dfrac{1}{2} bh$$:​

​​​$$\begin{aligned}A &= \dfrac{1}{2}bh \\\\&= \dfrac{1}{2} \times 2 \sqrt{10} \times 2 \sqrt{10} \\&=20 \end{aligned}$$​​

​​

Therefore, the area of the triangle is $$\underline{20 }$$.