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The circle C with centre T and radius r has equation $$x^2 + y^2 - 20x - 16y + 139 = 0$$ (a) Find the coordinates of the centre of C - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 3

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The circle C with centre T and radius r has equation $$x^2 + y^2 - 20x - 16y + 139 = 0$$ (a) Find the coordinates of the centre of C. (b) Show that r = 5. The li... show full transcript

Worked Solution & Example Answer:The circle C with centre T and radius r has equation $$x^2 + y^2 - 20x - 16y + 139 = 0$$ (a) Find the coordinates of the centre of C - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 3

Step 1

Find the coordinates of the centre of C.

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Answer

To find the centre of the circle represented by the equation

x2+y2−20x−16y+139=0x^2 + y^2 - 20x - 16y + 139 = 0

we can rewrite it in standard form. Let's group the x and y terms and complete the square.

  1. Rearrange the equation:
    x2−20x+y2−16y=−139x^2 - 20x + y^2 - 16y = -139

  2. Complete the square for x:
    x2−20x=(x−10)2−100x^2 - 20x = (x - 10)^2 - 100

  3. Complete the square for y:
    y2−16y=(y−8)2−64y^2 - 16y = (y - 8)^2 - 64

  4. Substitute back into the equation:
    (x−10)2−100+(y−8)2−64=−139(x - 10)^2 - 100 + (y - 8)^2 - 64 = -139

  5. Simplify:
    (x−10)2+(y−8)2=25(x - 10)^2 + (y - 8)^2 = 25

Thus, the centre T of the circle C is at the coordinates (10, 8).

Step 2

Show that r = 5.

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Answer

From the equation derived above:

(x−10)2+(y−8)2=25(x - 10)^2 + (y - 8)^2 = 25

The radius r is given by the square root of the right side of the equation:

r=extsqrt(25)=5.r = ext{sqrt}(25) = 5.
This confirms that r = 5.

Step 3

Find the y coordinate of P and the y coordinate of Q.

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Answer

Using the line L with equation x = 13, substitute x into the circle's equation:

  1. Substitute x = 13:
    (13−10)2+(y−8)2=25(13 - 10)^2 + (y - 8)^2 = 25
    which simplifies to
    32+(y−8)2=253^2 + (y - 8)^2 = 25
    or
    (y−8)2=16.(y - 8)^2 = 16.

  2. Taking the square root gives:
    y−8=ext±4.y - 8 = ext{±}4.
    Thus, the two solutions are:
    y=12extandy=4.y = 12 ext{ and } y = 4.

So, the coordinates are P(13, 12) and Q(13, 4).

Step 4

Find the perimeter of the sector PTQ.

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Answer

The angle PTQ is given as 1.855 radians, and we need to calculate the perimeter of the sector PTQ.

  1. The perimeter of the sector consists of the two radii and the arc length PQ.

  2. The arc length PQ can be calculated using the formula:
    extArcLength=rimesheta=5imes1.855=9.275.ext{Arc Length} = r imes heta = 5 imes 1.855 = 9.275.

  3. The total perimeter is given by:
    extPerimeter=2r+extArcLength=2(5)+9.275=10+9.275=19.275.ext{Perimeter} = 2r + ext{Arc Length} = 2(5) + 9.275 = 10 + 9.275 = 19.275.

Thus, the perimeter of the sector PTQ is 19.275.

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