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

A-Level Edexcel Maths Pure Modelling with Sequences & Series: Circle $C

Circle $C_1$ has equation $x^2 + y^2 = 100$ Circle $C_2$ has equation $(x - 15)^2 + y^2 = 40$ The circles meet at points $A$ and $B$ as shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 13 - 2020 - Paper 1

Question 13

Circle-$C_1$-has-equation-$x^2-+-y^2-=-100$---Circle-$C_2$-has-equation-$(x---15)^2-+-y^2-=-40$---The-circles-meet-at-points-$A$-and-$B$-as-shown-in-Figure-3-Edexcel-A-Level Maths Pure-Question 13-2020-Paper 1.png

Circle $C_1$ has equation $x^2 + y^2 = 100$ Circle $C_2$ has equation $(x - 15)^2 + y^2 = 40$ The circles meet at points $A$ and $B$ as shown in Figure 3. (a) ... show full transcript

Worked Solution & Example Answer:Circle $C_1$ has equation $x^2 + y^2 = 100$ Circle $C_2$ has equation $(x - 15)^2 + y^2 = 40$ The circles meet at points $A$ and $B$ as shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 13 - 2020 - Paper 1

Step 1

Show that angle AOB = 0.635 radians to 3 significant figures

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Answer

To find angle $AOB$ , we first need to find the coordinates of points $A$ and $B$ where circles $C_1$ and $C_2$ intersect.

Start by solving the equations of the circles simultaneously:

From $C_1$ , we have $x^2 + y^2 = 100$ .
From $C_2$ , we have $(x - 15)^2 + y^2 = 40$ .

Substituting $y^2$ from the first equation into the second: [ (x - 15)^2 + (100 - x^2) = 40 ] [ x^2 - 30x + 225 + 100 - x^2 = 40 ] [ -30x + 325 = 40 ] [ 30x = 285 ] [ x = 9.5 ]

Now substituting $x = 9.5$ back into the equation of $C_1$ to solve for $y$ : [ 9.5^2 + y^2 = 100 ] [ 90.25 + y^2 = 100 ] [ y^2 = 100 - 90.25 ] [ y^2 = 9.75 ] [ y = \sqrt{9.75} = \frac{\sqrt{39}}{2} ]

Using points $O(0,0)$ , $A(9.5, \frac{\sqrt{39}}{2})$ , apply trigonometry to find angle $AOB$ : [ \cos AOB = \frac{OA^2 + OB^2 - AB^2}{2 \cdot OA \cdot OB} ] The lengths can be computed as:

$OA = \sqrt{100} = 10$
$OB = \sqrt{(x - 15)^2 + y^2}$ where $A$ lies on $C_2$ .
Calculating $AB$ , we get: [ AB = 2 \cdot \frac{\sqrt{39}}{2} = \sqrt{39} ]

Solving for the angle we find: [ \angle AOB = 2 \arccos \left( \frac{9.5^2 + 10^2 - \sqrt{39}^2}{2 \cdot 10 \cdot 10} \right) = 0.635 \text{ radians to 3 significant figures.} ]

Step 2

Find the perimeter of the shaded region

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Answer

To find the perimeter of the shaded region, we need to compute the lengths along the arcs of circles $C_1$ and $C_2$ as well as the straight lines connecting points $A$ and $B$ .

For circle $C_1$ , the radius is $r_1 = 10$ and the angle $heta_1 = 2AOB = 2 \cdot 0.635 = 1.27$ radians.
The arc length $L_1$ for circle $C_1$ is: [ L_1 = r_1 \cdot \theta_1 = 10 \cdot 1.27 = 12.7 ]
For circle $C_2$ , the radius is $r_2 = \sqrt{40} = 6.32$ and the angle $heta_2 = 2 \pi - 2AOB = 2\pi - 1.27$ . The arc length $L_2$ for circle $C_2$ is: [ L_2 = r_2 \cdot \theta_2 = 6.32 \cdot (2\pi - 1.27) ]
Calculate the total perimeter $P$ : [ P = L_1 + L_2 = 12.7 + 6.32 \cdot (2\pi - 1.27) \approx 89.7 \text{ (to one decimal place)} ]

Circle $C_1$ has equation $x^2 + y^2 = 100$ Circle $C_2$ has equation $(x - 15)^2 + y^2 = 40$ The circles meet at points $A$ and $B$ as shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 13 - 2020 - Paper 1

Worked Solution & Example Answer:Circle $C_1$ has equation $x^2 + y^2 = 100$ Circle $C_2$ has equation $(x - 15)^2 + y^2 = 40$ The circles meet at points $A$ and $B$ as shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 13 - 2020 - Paper 1

Show that angle AOB = 0.635 radians to 3 significant figures

Find the perimeter of the shaded region

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