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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 12 - 2020 - Paper 1

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Question 12

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 12-2020-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. (... 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 12 - 2020 - Paper 1

Step 1

Show that angle AOB = 0.635 radians to 3 significant figures

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Answer

To find angle AOBAOB, begin by determining the coordinates of the intersection points of the circles C1C_1 and C2C_2.

  1. Find Coordinates of Points A and B:

    • Solve the system of equations:

      x2+y2=100ag1x^2 + y^2 = 100 ag{1} (x15)2+y2=40ag2(x - 15)^2 + y^2 = 40 ag{2}
    • Expanding (2):

      x230x+225+y2=40x^2 - 30x + 225 + y^2 = 40
    • Substitute (1) into this to find:

      10030x+225=40100 - 30x + 225 = 40
    • This simplifies to:

      30x+325=40 30x=285 x=9.5-30x + 325 = 40 \ -30x = -285 \ x = 9.5
    • Substitute x=9.5x = 9.5 back into (1) to find yy:

      (9.5)^2 + y^2 = 100 \ 90.25 + y^2 = 100 \ y^2 = 9.75 \ y = rac{ ext{sqrt}(39)}{2}
  2. Using Trigonometry to Find Angle AOB:

    • In triangle AOBAOB, using the coordinates for points AA and BB with center O(0,0)O(0, 0):

    • Use cosine rule or derive directly using tangent:

      ext{cos} heta = rac{OA^2 + OB^2 - AB^2}{2 imes OA imes OB}
    • Solve for the angle in radians:

      heta = 2 ext{arccos}igg( rac{9.5}{10}igg) = 0.635
    • Angle AOBAOB is thus confirmed as 0.6350.635 radians.

Step 2

Find the perimeter of the shaded region

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Answer

To find the perimeter of the shaded region bounded by circles C1C_1 and C2C_2, follow these steps:

  1. Determine the Radii:

    • Circle C1C_1: Radius r1=10r_1 = 10 (from x2+y2=100x^2 + y^2 = 100)
    • Circle C2C_2: Radius r2=extsqrt(40)=6.32r_2 = ext{sqrt}(40) = 6.32 (from (x15)2+y2=40(x - 15)^2 + y^2 = 40)
  2. Calculate the Length of Arc OA and Arc OB:

    • The angle AOB=0.635AOB = 0.635 radians, so the lengths are:

      L1=r1heta=10imes0.635=6.35L_1 = r_1 heta = 10 imes 0.635 = 6.35 L2=r2heta=6.32imes0.635=4.01L_2 = r_2 heta = 6.32 imes 0.635 = 4.01
  3. Perimeter Calculation:

    • The total perimeter is the sum of arcs OA and OB:

      P=L1+L2=6.35+4.01=10.36P = L_1 + L_2 = 6.35 + 4.01 = 10.36
    • Round to one decimal place gives:

      Pext(shadedperimeter)=10.4P ext{ (shaded perimeter)} = 10.4

Thus, the perimeter of the shaded region is approximately 10.4 units.

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