| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Find equations of tangent lines with given gradient or from external point using discriminant |
| Difficulty | Standard +0.3 This is a standard two-part circle question requiring substitution to find intersection points, distance formula, and using the perpendicular distance from center to tangent equals radius. Part (a) involves routine algebraic manipulation with surds, while part (b) requires applying the tangent condition formula—both are well-practiced techniques with no novel insight needed, making it slightly easier than average. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^2 + (2x+5)^2 = 20\) leading to \(x^2 + 4x^2 + 20x + 25 = 20\) | M1 | Substitute \(y = 2x + 5\) and expand bracket |
| \((5)(x^2 + 4x + 1)[= 0]\) | A1 | 3-term quadratic |
| \(x = \frac{-4 \pm \sqrt{16-4}}{2}\) | M1 | OE. Apply formula or complete the square |
| \(A = \left(-2+\sqrt{3}, 1+2\sqrt{3}\right)\) | A1 | Or 2 correct \(x\) values |
| \(B = \left(-2-\sqrt{3}, 1-2\sqrt{3}\right)\) | A1 | Or all values correct. SC B1 all 4 values correct in surd form without working. SC B1 all 4 values correct in decimal form from correct formula or completion of the square |
| \(AB^2 = \textit{their}(x_2-x_1)^2 + \textit{their}(y_2-y_1)^2\) | M1 | Using *their* coordinates in a correct distance formula. Condone one sign error in \(x_2-x_1\) or \(y_2-y_1\) |
| \(\left[AB^2 = 48 + 12\right.\) leading to\(\left.\right] AB = \sqrt{60}\) | A1 | OE. CAO. Do not accept decimal answer. Answer must come from use of surd form in distance formula |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^2 + m^2(x-10)^2 = 20\) | *M1 | Finding equation of tangent and substituting into circle equation |
| \(x^2(m^2+1) - 20m^2x + 20(5m^2-1)[=0]\) | DM1 | OE. Brackets expanded and all terms collected on one side |
| \([b^2 - 4ac =] 400m^4 - 80(m^2+1)(5m^2-1)\) | M1 | Using correct coefficients from *their* quadratic equation |
| \(400m^4 - 80(5m^4 + 4m^2 - 1) = 0 \to (-80)(4m^2-1) = 0\) | A1 | OE. Must have '\(=0\)' for A1 |
| \(m = \pm\frac{1}{2}\) | A1 | |
| Alternative method: Length \(l\) of tangent given by \(l^2 = 10^2 - 20\) | M1 | |
| \(l = \sqrt{80}\) | A1 | |
| \(\tan\alpha = \frac{\sqrt{20}}{\sqrt{80}} = \frac{1}{2}\) | M1 A1 | Where \(\alpha\) is the angle between the tangent and the \(x\)-axis |
| \(m = \pm\frac{1}{2}\) | A1 |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 + (2x+5)^2 = 20$ leading to $x^2 + 4x^2 + 20x + 25 = 20$ | M1 | Substitute $y = 2x + 5$ and expand bracket |
| $(5)(x^2 + 4x + 1)[= 0]$ | A1 | 3-term quadratic |
| $x = \frac{-4 \pm \sqrt{16-4}}{2}$ | M1 | OE. Apply formula or complete the square |
| $A = \left(-2+\sqrt{3}, 1+2\sqrt{3}\right)$ | A1 | Or 2 correct $x$ values |
| $B = \left(-2-\sqrt{3}, 1-2\sqrt{3}\right)$ | A1 | Or all values correct. SC B1 all 4 values correct in surd form without working. SC B1 all 4 values correct in decimal form from correct formula or completion of the square |
| $AB^2 = \textit{their}(x_2-x_1)^2 + \textit{their}(y_2-y_1)^2$ | M1 | Using *their* coordinates in a correct distance formula. Condone one sign error in $x_2-x_1$ or $y_2-y_1$ |
| $\left[AB^2 = 48 + 12\right.$ leading to$\left.\right] AB = \sqrt{60}$ | A1 | OE. CAO. Do not accept decimal answer. Answer must come from use of surd form in distance formula |
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## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 + m^2(x-10)^2 = 20$ | *M1 | Finding equation of tangent and substituting into circle equation |
| $x^2(m^2+1) - 20m^2x + 20(5m^2-1)[=0]$ | DM1 | OE. Brackets expanded and all terms collected on one side |
| $[b^2 - 4ac =] 400m^4 - 80(m^2+1)(5m^2-1)$ | M1 | Using correct coefficients from *their* quadratic equation |
| $400m^4 - 80(5m^4 + 4m^2 - 1) = 0 \to (-80)(4m^2-1) = 0$ | A1 | OE. Must have '$=0$' for A1 |
| $m = \pm\frac{1}{2}$ | A1 | |
| **Alternative method:** Length $l$ of tangent given by $l^2 = 10^2 - 20$ | M1 | |
| $l = \sqrt{80}$ | A1 | |
| $\tan\alpha = \frac{\sqrt{20}}{\sqrt{80}} = \frac{1}{2}$ | M1 A1 | Where $\alpha$ is the angle between the tangent and the $x$-axis |
| $m = \pm\frac{1}{2}$ | A1 | |
---9 The line $y = 2 x + 5$ intersects the circle with equation $x ^ { 2 } + y ^ { 2 } = 20$ at $A$ and $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $A$ and $B$ in surd form and hence find the exact length of the chord $A B$.\\
A straight line through the point $( 10,0 )$ with gradient $m$ is a tangent to the circle.
\item Find the two possible values of $m$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q9 [12]}}