Circle $C_1$ has equation $x^2 + y^2 = 100$\nCircle $C_2$ has equation $(x - 15)^2 + y^2 = 40$\nThe circles meet at points $A$ and $B$ as shown in Figure 3.\n\n(a) Show that angle $AOB = 0.635$ radians to 3 significant figures, where $O$ is the origin.\n(b) The region shown shaded in Figure 3 is bounded by $C_1$ and $C_2$\n\n(b) Find the perimeter of the shaded region, giving your answer to one decimal place. - Edexcel - A-Level Maths Pure - Question 13 - 2020 - Paper 1

To find angle $AOB$, we first need the coordinates of the intersection points of the circles.\n1. Set the equations of the circles equal to find $x$ and $y$: \n
From $C_1$: $y^2 = 100 - x^2$\n From $C_2$: $(x - 15)^2 + y^2 = 40$\n
Substitute $y^2$: $(x - 15)^2 + (100 - x^2) = 40$\n
Expanding gives: $x^2 - 30x + 225 + 100 - x^2 = 40$. \n So, we solve \n [ -30x + 325 = 40 ] \n [ -30x = -285 ] \n [ x = 9.5 ] \n
Substituting back, \n [ y^2 = 100 - (9.5)^2 = 100 - 90.25 = 9.75 ] \n Thus y = rac{- adical{39}}{2} or $y ext{ and from } C_1$: this gives coordinates of intersection points as \n A(9.5, rac{ adical{39}}{2}) and B(9.5, -rac{ adical{39}}{2}).\n2. With triangle $AOB$, use trigonometry: consider cosine rule\n We know $OA = OB = 10$ (radius of $C_1$) and $AB = adical{39}$. \n [ ext{Thus, cos}(AOB) = \frac{OA^2 + OB^2 - AB^2}{2 imes OA imes OB} ] \n [ = \frac{10^2 + 10^2 - (\radical{39})^2}{2 \times 10 \times 10} ] \n [ = \frac{200 - 39}{200} = \frac{161}{200} ] \n Hence, \n [ angle AOB = 2 \cdot \arccos\left(\frac{161}{200}\right) \approx 0.635\text{ radians} ]