| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Standard +0.3 Part (a) requires verifying tangency by checking the radius is perpendicular to the line (gradient product = -1) and that A lies on the circle (distance = radius), which are standard techniques. Part (b) involves finding another circle with the same tangent at A, requiring understanding that the centre lies on the normal through A at distance √52 in the opposite direction—a slightly less routine application but still follows directly from tangent properties. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((5-1)^2 + (11-5)^2 = 52\) or \(\frac{11-5}{5-1}\) | M1 | For substituting \((1,5)\) into circle equation or showing gradient \(= \frac{3}{2}\) |
| For both circle equation and gradient, proving line is perpendicular and stating that \(A\) lies on the circle | A1 | Clear reasoning |
| Alternative: \((x-5)^2 + (y-11)^2 = 52\) and \(y-5 = -\frac{2}{3}(x-1)\) | M1 | Both equations seen and attempt to solve. May see \(y = -\frac{2}{3}x + \frac{17}{3}\) |
| Solving simultaneously to obtain \((y-5)^2 = 0\) or \((x-1)^2 = 0 \Rightarrow 1\) root or tangent or discriminant \(= 0 \Rightarrow 1\) root or tangent | A1 | Clear reasoning |
| Alternative: \(\frac{dy}{dx} = \frac{10-2x}{2y-22} = \frac{10-2}{10-22}\) | M1 | Attempting implicit differentiation of circle equation and substitute \(x=1\) and \(y=5\) |
| Showing gradient of circle at \(A\) is \(-\frac{2}{3}\) | A1 | Clear reasoning |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Centre is \((-3, -1)\) | B1 B1 | B1 for each correct co-ordinate |
| Equation is \((x+3)^2 + (y+1)^2 = 52\) | B1 FT | FT *their* centre, but not if either \((1,5)\) or \((5,11)\). Do not accept \(\sqrt{52^2}\) |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(5-1)^2 + (11-5)^2 = 52$ or $\frac{11-5}{5-1}$ | M1 | For substituting $(1,5)$ into circle equation or showing gradient $= \frac{3}{2}$ |
| For both circle equation and gradient, proving line is perpendicular and stating that $A$ lies on the circle | A1 | Clear reasoning |
| **Alternative:** $(x-5)^2 + (y-11)^2 = 52$ and $y-5 = -\frac{2}{3}(x-1)$ | M1 | Both equations seen and attempt to solve. May see $y = -\frac{2}{3}x + \frac{17}{3}$ |
| Solving simultaneously to obtain $(y-5)^2 = 0$ or $(x-1)^2 = 0 \Rightarrow 1$ root or tangent or discriminant $= 0 \Rightarrow 1$ root or tangent | A1 | Clear reasoning |
| **Alternative:** $\frac{dy}{dx} = \frac{10-2x}{2y-22} = \frac{10-2}{10-22}$ | M1 | Attempting implicit differentiation of circle equation and substitute $x=1$ and $y=5$ |
| Showing gradient of circle at $A$ is $-\frac{2}{3}$ | A1 | Clear reasoning |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Centre is $(-3, -1)$ | B1 B1 | B1 for each correct co-ordinate |
| Equation is $(x+3)^2 + (y+1)^2 = 52$ | B1 FT | FT *their* centre, but not if either $(1,5)$ or $(5,11)$. Do not accept $\sqrt{52^2}$ |7 The point $A$ has coordinates $( 1,5 )$ and the line $l$ has gradient $- \frac { 2 } { 3 }$ and passes through $A$. A circle has centre $( 5,11 )$ and radius $\sqrt { 52 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $l$ is the tangent to the circle at $A$.
\item Find the equation of the other circle of radius $\sqrt { 52 }$ for which $l$ is also the tangent at $A$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q7 [5]}}