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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 4 - 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 lin... 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 4 - 2012 - Paper 3

Step 1

Find the coordinates of the centre of C.

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Answer

To find the coordinates of the center of the circle given by the equation

x2+y2−20x−16y+139=0,x^2 + y^2 - 20x - 16y + 139 = 0,
we can rearrange it into the standard form of a circle equation, which is

(x−h)2+(y−k)2=r2,(x - h)^2 + (y - k)^2 = r^2,
where (h,k)(h, k) is the center of the circle.

We complete the square for both x and y:

  • For xx:

    1. Take the coefficient of xx, which is -20, halve it to -10, and square it to get 100.
    2. Adjust the equation accordingly: ( (x^2 - 20x + 100) \rightarrow (x - 10)^2 ).

  • For yy:

    1. Take the coefficient of yy, which is -16, halve it to -8, and square it to get 64.
    2. Adjust the equation accordingly: ( (y^2 - 16y + 64) \rightarrow (y - 8)^2 ).

Now rewrite the circle’s equation:

(x−10)2+(y−8)2=100+64−139(x - 10)^2 + (y - 8)^2 = 100 + 64 - 139
(x−10)2+(y−8)2=25(x - 10)^2 + (y - 8)^2 = 25
This reveals that the center of the circle C is at (10,8)(10, 8).

Step 2

Show that r = 5

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Answer

From the completed square form, we derived that:

r2=25r^2 = 25
Taking the square root, we find:

r=5.r = 5.
Thus, we have shown that the radius r is indeed 5.

Step 3

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

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Answer

To find the y-coordinates of points P and Q where the line x=13x = 13 intersects the circle, we substitute x=13x = 13 into the equation of the circle:

(13−10)2+(y−8)2=25(13 - 10)^2 + (y - 8)^2 = 25
This simplifies to:

32+(y−8)2=253^2 + (y - 8)^2 = 25
9+(y−8)2=259 + (y - 8)^2 = 25
Subtracting 9 from both sides gives:

(y−8)2=16.(y - 8)^2 = 16.
Taking square roots, we derive:

y−8=4y - 8 = 4
or
y−8=−4y - 8 = -4
Thus: y=12y = 12
or
y=4.y = 4.
Therefore, the coordinates of P are (13, 12) and the coordinates of Q are (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 the radius r is 5. The perimeter of the sector PTQ consists of the arc PQ and the two radii PT and QT.

Arc length = r⋅θ=5⋅1.855=9.275r \cdot \theta = 5 \cdot 1.855 = 9.275
The total perimeter is calculated by adding the lengths of the two radii:

Perimeter=9.275+5+5=19.275.\text{Perimeter} = 9.275 + 5 + 5 = 19.275.
Thus, the perimeter of the sector PTQ is approximately 19.275.

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