Standard +0.3 This is a standard circle geometry problem requiring completion of the square to find the center, substituting x=0 to find points A and B, then using the property that tangents from an external point have equal length or that radius is perpendicular to tangent. While it involves multiple steps (4-5 marks worth), each step uses routine techniques taught in P1 with no novel insight required, making it slightly easier than average.
8 A circle with equation \(x ^ { 2 } + y ^ { 2 } - 6 x + 2 y - 15 = 0\) meets the \(y\)-axis at the points \(A\) and \(B\). The tangents to the circle at \(A\) and \(B\) meet at the point \(P\).
Find the coordinates of \(P\).
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\includegraphics[max width=\textwidth, alt={}, center]{d6976a4b-aecf-43f1-a3f2-bcad37d03585-10_2723_37_136_2010}
Substitute \(x = 0\) and attempt solution of 3-term quadratic equation in \(y\)
M1
If \(y = 0\) used can score a maximum of M0 A0 B1 M1 A0 A1FT DM1 A0, i.e. 4/8
\(-5\) and \(3\)
A1
B1 SC if no working to solve the quadratic
State or imply centre of circle is \((3, -1)\)
B1
Condone errors which don't affect finding centre. May be implied by the correct final \(y\) coordinate
Attempt gradient of \(AC\) or \(BC\)
*M1
\(-\frac{4}{3}\) or \(\frac{4}{3}\)
A1
State or imply gradient of tangent is \(\frac{3}{4}\) or \(-\frac{3}{4}\)
A1FT
Following *their* gradient of radius. Only FT when previous 2 marks are M1 A0
Either solve simultaneous equations (of 2 tangent equations) to find \(x\)-coordinate, Or substitute \(y\)-value of centre into either tangent equation
DM1
\(x = -\frac{16}{3}\), \(y = -1\)
A1
Alternative Method 1: for the 4th and 5th marks
Answer
Marks
Guidance
Answer
Mark
Guidance
Rearrange and differentiate the circle equation or differentiate implicitly
(M1)
Replaces the second M1
\(\frac{dy}{dx} = \frac{3-x}{y+1}\) or \(\frac{dy}{dx} = \frac{3-x}{\left(25-(x-3)^2\right)^{\frac{1}{2}}}\)
(A1)
Replaces the second A1
Alternative Method 2: for the last 5 marks
Answer
Marks
Guidance
Answer
Mark
Guidance
\(\widehat{ACP} = \widehat{MAP} = \tan^{-1}\frac{4}{3}\) or identifying similar triangles \(PMA\) and \(AMC\)
(M1A1)
\(C\) is the circle centre, \(P\) is intersection of the two tangents, \(M\) is intersection of \(PC\) and the \(y\)-axis
\(\tan MAP = \frac{PM}{4} \cdot \frac{4}{3} = \frac{PM}{4}\), \(PM = \frac{16}{3}\) or use of similar triangles
(M1A1)
\(P\) is \(\left(\frac{-16}{3}, -1\right)\)
(A1)
Alternative Method 3: for the last 5 marks
Answer
Marks
Guidance
Answer
Mark
Guidance
Pythagoras on triangle \(PAC\), \(PC^2 = PA^2 + AC^2\)
(M1)
Identifies the required 3 sides and sets up formula
## Question 8:
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $x = 0$ and attempt solution of 3-term quadratic equation in $y$ | M1 | If $y = 0$ used can score a maximum of M0 A0 B1 M1 A0 A1FT DM1 A0, i.e. 4/8 |
| $-5$ and $3$ | A1 | **B1 SC** if no working to solve the quadratic |
| State or imply centre of circle is $(3, -1)$ | B1 | Condone errors which don't affect finding centre. May be implied by the correct final $y$ coordinate |
| Attempt gradient of $AC$ or $BC$ | *M1 | |
| $-\frac{4}{3}$ or $\frac{4}{3}$ | A1 | |
| State or imply gradient of tangent is $\frac{3}{4}$ or $-\frac{3}{4}$ | A1FT | Following *their* gradient of radius. Only FT when previous 2 marks are M1 A0 |
| Either solve simultaneous equations (of 2 tangent equations) to find $x$-coordinate, Or substitute $y$-value of centre into either tangent equation | DM1 | |
| $x = -\frac{16}{3}$, $y = -1$ | A1 | |
**Alternative Method 1: for the 4th and 5th marks**
| Answer | Mark | Guidance |
|--------|------|----------|
| Rearrange and differentiate the circle equation or differentiate implicitly | (M1) | Replaces the second M1 |
| $\frac{dy}{dx} = \frac{3-x}{y+1}$ or $\frac{dy}{dx} = \frac{3-x}{\left(25-(x-3)^2\right)^{\frac{1}{2}}}$ | (A1) | Replaces the second A1 |
**Alternative Method 2: for the last 5 marks**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\widehat{ACP} = \widehat{MAP} = \tan^{-1}\frac{4}{3}$ or identifying similar triangles $PMA$ and $AMC$ | (M1A1) | $C$ is the circle centre, $P$ is intersection of the two tangents, $M$ is intersection of $PC$ and the $y$-axis |
| $\tan MAP = \frac{PM}{4} \cdot \frac{4}{3} = \frac{PM}{4}$, $PM = \frac{16}{3}$ or use of similar triangles | (M1A1) | |
| $P$ is $\left(\frac{-16}{3}, -1\right)$ | (A1) | |
**Alternative Method 3: for the last 5 marks**
| Answer | Mark | Guidance |
|--------|------|----------|
| Pythagoras on triangle $PAC$, $PC^2 = PA^2 + AC^2$ | (M1) | Identifies the required 3 sides and sets up formula |
| $PC^2 = (PM+3)^2$, $PA^2 = PM^2 + 4^2$, $AC = \text{radius} = 5$ | (A1) | Finds each side with two in terms of $PM$ |
| $(PM+3)^2 = PM^2 + 4^2 + 5^2$ leads to $6PM = 32$, $PM = \frac{16}{3}$ | (M1A1) | Sets up and solves equation |
| $P$ is $\left(\frac{-16}{3}, -1\right)$ | (A1) | |
| **Total** | **8** | |
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8 A circle with equation $x ^ { 2 } + y ^ { 2 } - 6 x + 2 y - 15 = 0$ meets the $y$-axis at the points $A$ and $B$. The tangents to the circle at $A$ and $B$ meet at the point $P$.
Find the coordinates of $P$.\\
\includegraphics[max width=\textwidth, alt={}, center]{d6976a4b-aecf-43f1-a3f2-bcad37d03585-10_71_1659_466_244}\\
\includegraphics[max width=\textwidth, alt={}, center]{d6976a4b-aecf-43f1-a3f2-bcad37d03585-10_2723_37_136_2010}\\
\hfill \mbox{\textit{CAIE P1 2024 Q8 [8]}}