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 $S$ .

Maths; Circles; KS5 Year 12; Intersection of two circles

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 $RBS$ .

Maths; Circles; KS5 Year 12; Intersection of two circles

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 $\underline{20 }$ .