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The centre of a circle is the point with coordinates (−1, 3) The point A with coordinates (6, 8) lies on the circle - Edexcel - GCSE Maths - Question 21 - 2022 - Paper 1

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The centre of a circle is the point with coordinates (−1, 3) The point A with coordinates (6, 8) lies on the circle. Find an equation of the tangent to the circle ... show full transcript

Worked Solution & Example Answer:The centre of a circle is the point with coordinates (−1, 3) The point A with coordinates (6, 8) lies on the circle - Edexcel - GCSE Maths - Question 21 - 2022 - Paper 1

Step 1

Find the gradient of the line from the centre of the circle to the point (6, 8)

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Answer

To find the gradient (m) of the line joining the centre of the circle (−1, 3) to point A (6, 8), we can use the formula:

m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}

In this case, substituting the coordinates: m=836(1)=57m = \frac{8 - 3}{6 - (-1)} = \frac{5}{7}

Step 2

Find the negative reciprocal of the gradient for the tangent line

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Answer

The gradient of the tangent line (m_t) is the negative reciprocal of the gradient calculated above:

mt=1m=157=75m_t = -\frac{1}{m} = -\frac{1}{\frac{5}{7}} = -\frac{7}{5}

Step 3

Use point-slope form to find the equation of the tangent line

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Answer

Now using the point-slope form of the line equation, which is: yy1=mt(xx1)y - y_1 = m_t (x - x_1) Given point A (6, 8):

y8=75(x6)y - 8 = -\frac{7}{5}(x - 6)

Simplifying this:

  1. Distributing: y8=75x+425y - 8 = -\frac{7}{5} x + \frac{42}{5}
  2. Rearranging to standard form: y=75x+425+8y = -\frac{7}{5} x + \frac{42}{5} + 8 y=75x+425+405y = -\frac{7}{5} x + \frac{42}{5} + \frac{40}{5} y=75x+825y = -\frac{7}{5} x + \frac{82}{5}
  3. Multiplying everything by 5 to eliminate the fraction: 5y=7x+825y = -7x + 82 7x+5y82=07x + 5y - 82 = 0

Step 4

Final result in the form $ax + by + c = 0$

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Answer

Thus, the equation of the tangent line is:

7x+5y82=07x + 5y - 82 = 0

Where a = 7, b = 5, and c = -82, fulfilling the requirement that a, b, and c are integers.

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