Higher June 2022 Paper 1 Q20
20 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 at \(A\).
Give your answer in the form \(ax + by + c = 0\) where \(a\), \(b\) and \(c\) are integers. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(7x + 5y - 82 = 0\) | P1 | for process to work out the gradient of the line from the centre of the circle to the point (6,8) eg \(\dfrac{8 - 3}{6 - -1}\ \left(= \dfrac{5}{7}\right)\) |
| P1 | (dep P1) for using \(mn = -1\) eg \(-1 \div \text{``}\dfrac{5}{7}\text{''}\ \left(= -\dfrac{7}{5}\right)\) | |
| P1 | for substituting \((6, 8)\) into \(y = \text{``}{-}\dfrac{7}{5}\text{''}x + c\) or for \((y - 8) = \text{``}{-}\dfrac{7}{5}\text{''}(x - 6)\) or for \(y = -\dfrac{7}{5}x + \dfrac{82}{5}\) oe | |
| A1 | \(7x + 5y - 82 = 0\) oe SC B2 for answer of \(5x + 7y - 86 = 0\) oe in any form |
Additional guidance
Must be in form \(ax + by + c = 0\) with integer coefficients, eg \(82 - 7x - 5y = 0\)