Higher November 2024 Paper 1 Q17
17
(a) A circle has centre (0, 0) and circumference \(36\pi\)
Work out the equation of the circle. [2 marks]
(b) Point \(J\) has coordinates (15, 0) and point \(K\) has coordinates (30, −5)
Work out the equation of the straight line through \(J\) and \(K\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((r =)\ 36\pi \div 2\pi\) or \((r =)\ 36 \div 2\) or \((r =)\ 18\) or \((36\pi \div 2\pi)^2\) or \((36 \div 2)^2\) or \(18^2\) or 324 | M1 | |
| \(x^2 + y^2 = 18^2\) or \(x^2 + y^2 = 324\) | A1 |
Additional guidance
| \(x^2 + y^2 = 18\) or \(a^2 + b^2 = 18\) | M1A0 |
| \(a^2 + b^2 = 18^2\) or \(a^2 + b^2 = 324\) | M1A0 |
| Allow \(x^2 + y^2 = 18^2\) followed by an incorrect evaluation of \(18^2\) eg \(x^2 + y^2 = 18^2\), \(18 \times 18 = 506\), \(x^2 + y^2 = 506\) | M1A1 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: uses the gradient with either point to work out the intercept | ||
| \(\dfrac{0 - (-5)}{15 - 30}\) or \(-\dfrac{1}{3}\) | M1 | oe |
| their \(-\dfrac{1}{3} \times 15 + c = 0\) or their \(-\dfrac{1}{3} \times 30 + c = -5\) | M1dep | oe |
| \((m =)\ -\dfrac{1}{3}\) and \((c =)\ 0 + \dfrac{1}{3} \times 15\) or \((m =)\ -\dfrac{1}{3}\) and \((c =)\ -5 + \dfrac{1}{3} \times 30\) or \((m =)\ -\dfrac{1}{3}\) and \((c =)\ 5\) | M1dep | oe equation in \(c\) or expression for \(c\) |
| \(y = -\dfrac{1}{3}x + 5\) or \(x + 3y - 15 = 0\) | A1 | oe equation with terms collected |
| Alternative method 2: uses the gradient with \(y - y_1 = m(x - x_1)\) | ||
| \(\dfrac{0 - (-5)}{15 - 30}\) or \(-\dfrac{1}{3}\) | M1 | oe |
| \(y - 0 =\) their \(-\dfrac{1}{3}(x - 15)\) or \(y - (-5) =\) their \(-\dfrac{1}{3}(x - 30)\) | M1dep | oe |
| \(y = -\dfrac{1}{3}x + 5\) or \(x + 3y - 15 = 0\) | A2 | oe equation with terms collected A1 \((m =)\ -\dfrac{1}{3}\) and \((c =)\ 5\) |
| Alternative method 3: finds gradient and \(y\)-intercept separately | ||
| \(\dfrac{0 - (-5)}{15 - 30}\) or \(-\dfrac{1}{3}\) | M1 | oe |
| (\(y\)-intercept =) (0, 5) or \(c = 5\) | M1 | |
| \(y = -\dfrac{1}{3}x + 5\) or \(x + 3y - 15 = 0\) | A2 | oe equation with terms collected A1 \((m =)\ -\dfrac{1}{3}\) and \((c =)\ 5\) |
| Alternative method 4: simultaneous equations using both points | ||
| Correct elimination of one variable | M1 | eg \(0 = 15m + c\) and \(-5 = 30m + c\) and \(15m - 30m = 0 - (-5)\) or \(-15m = 5\) or \(m = -\dfrac{1}{3}\) |
| Correct substitution into a correct equation | M1dep | eg \(0 = 15 \times -\dfrac{1}{3} + c\) |
| \(y = -\dfrac{1}{3}x + 5\) or \(x + 3y - 15 = 0\) | A2 | oe equation with terms collected A1 \((m =)\ -\dfrac{1}{3}\) and \((c =)\ 5\) |
Additional guidance
| Allow an equivalent fraction for \(-\dfrac{1}{3}\) throughout | |
| As a decimal, allow \(-0.33\) or better for \(-\dfrac{1}{3}\) for the method marks, but correct recurring notation or at least \(-0.33..\) is needed for the A mark | |
| In alts 1 and 2, do not allow use of the negative inverse of their gradient | M1 max |