Higher June 2018 Paper 3 Q28
28 \(P\) is a point on the circle with equation \(\quad x^2 + y^2 = 80\)
\(P\) has \(x\)-coordinate 4 and is below the \(x\)-axis.
Not drawn accurately
Work out the equation of the tangent to the circle at \(P\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(4^2 + y^2 = 80\) or \(y = \sqrt{64}\) | M1 | oe May be implied from 8 on diagram |
| \(y = -8\) | A1 | Accept \((4, -8)\) |
| \(\dfrac{\text{their } -8}{4}\) or \(-2\) | M1 | oe gradient of radius \(OP\) |
| \(-1 \div\) their \(-2\) or \(\dfrac{1}{2}\) or \(-1 \div\) their gradient | M1 | gradient of tangent at \(P\) |
| \(y = \dfrac{1}{2}x - 10\) or \(\;y + 8 = \dfrac{1}{2}(x - 4)\) | A1 | oe Ignore further working |
Additional guidance
| \(y + 8 = \dfrac{1}{2}(x - 4)\) followed by error expanding and/or collecting terms | M1A1M1M1A1 |
| \(y = \dfrac{1}{2}x - 10\) in working and \(\dfrac{1}{2}x - 10\) only on answer | M1A1M1M1A1 |
| \(\dfrac{1}{2}x - 10\) | M1A1M1M1A0 |
| (\(y = \sqrt{64}\)) \(y = 8\) Gradient \(OP = 2\) Perpendicular gradient \(= -\dfrac{1}{2}\) | M1 A0 M1 M1 A0 |