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The circle C has radius 5 and touches the y-axis at the point (0, 9), as shown in Figure 4 - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 3

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The circle C has radius 5 and touches the y-axis at the point (0, 9), as shown in Figure 4. (a) Write down an equation for the circle C, that is shown in Figure 4. ... show full transcript

Worked Solution & Example Answer:The circle C has radius 5 and touches the y-axis at the point (0, 9), as shown in Figure 4 - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 3

Step 1

Write down an equation for the circle C

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Answer

To find the equation of the circle C, we need the center and the radius. The center of the circle is at (5, 9) since it is 5 units away from the y-axis and touches it at (0, 9) with a radius of 5. The general form of the equation for a circle is:

(xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2 where ( (h, k) ) is the center and ( r ) is the radius. Thus, substituting the values,

(x5)2+(y9)2=52(x - 5)^2 + (y - 9)^2 = 5^2 which simplifies to:

(x5)2+(y9)2=25(x - 5)^2 + (y - 9)^2 = 25

Step 2

Find the length of PT

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Answer

To find the length of PT, we first need to determine the coordinates of point T where the tangent from P(8, -7) touches the circle C. Since PT is perpendicular to the radius CT, we find the coordinates of C which is (5, 9).

  1. Calculate the distance PC:

    PC=(85)2+(79)2=32+(16)2=9+256=265PC = \sqrt{(8 - 5)^2 + (-7 - 9)^2} = \sqrt{3^2 + (-16)^2} = \sqrt{9 + 256} = \sqrt{265}

  2. Next, we use the relationship between the radius, PT, and PC:

    PT2+CT2=PC2PT^2 + CT^2 = PC^2 where (CT = 5), hence:

    PT2+52=265PT^2 + 5^2 = 265

    PT2+25=265PT^2 + 25 = 265

    PT2=26525PT^2 = 265 - 25

    PT2=240PT^2 = 240

  3. Finally, calculate PT:

    PT=240=1615=415PT = \sqrt{240} = \sqrt{16 \cdot 15} = 4\sqrt{15}

Thus, the length of PT is (4\sqrt{15} \approx 15.49).

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