OCR C1 — Question 7 11 marks

Exam BoardOCR
ModuleC1 (Core Mathematics 1)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeTangent equation at a known point on circle
DifficultyModerate -0.8 This is a straightforward C1 circles question requiring standard techniques: finding circle equation from diameter endpoints (midpoint for centre, distance formula for radius), finding x-intercepts by substituting y=0, and using perpendicular gradient for tangent. All steps are routine applications of formulas with no problem-solving insight needed, making it easier than average but not trivial due to the multi-step nature and 11 marks total.
Spec1.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.07m Tangents and normals: gradient and equations

The point \(A\) has coordinates \((4, 6)\). Given that \(OA\), where \(O\) is the origin, is a diameter of circle \(C\),
  1. find an equation for \(C\). [4]
Circle \(C\) crosses the \(x\)-axis at \(O\) and at the point \(B\). \begin{enumerate}[label=(\roman*)] \setcounter{enumi}{1} \item Find the coordinates of \(B\). [2] \item Find an equation for the tangent to \(C\) at \(B\), giving your answer in the form \(ax + by = c\), where \(a\), \(b\) and \(c\) are integers. [5]
AnswerMarks
(i) centre = \((2, 3)\)B1
radius = \(\sqrt{4+9} = \sqrt{13}\)M1
\(\therefore (x-2)^2 + (y-3)^2 = (\sqrt{13})^2\)M1
\((x-2)^2 + (y-3)^2 = 13\)A1
(ii) \(y = 0 \therefore (x-2)^2 + 9 = 13\)M1
\(x = 2 \pm \sqrt{4} = 0\) (at O) or \(4 \therefore B(4, 0)\)A1
(iii) grad of radius = \(\frac{0-3}{4-2} = -\frac{3}{2}\)M1
\(\therefore\) grad of tangent = \(-1 \div (-\frac{3}{2}) = \frac{2}{3}\)M1, A1
\(\therefore y - 0 = \frac{2}{3}(x-4)\)M1
\(3y = 2x - 8\)A1
\(2x - 3y = 8\)(11)
**(i)** centre = $(2, 3)$ | B1 |
radius = $\sqrt{4+9} = \sqrt{13}$ | M1 |
$\therefore (x-2)^2 + (y-3)^2 = (\sqrt{13})^2$ | M1 |
$(x-2)^2 + (y-3)^2 = 13$ | A1 |

**(ii)** $y = 0 \therefore (x-2)^2 + 9 = 13$ | M1 |
$x = 2 \pm \sqrt{4} = 0$ (at O) or $4 \therefore B(4, 0)$ | A1 |

**(iii)** grad of radius = $\frac{0-3}{4-2} = -\frac{3}{2}$ | M1 |
$\therefore$ grad of tangent = $-1 \div (-\frac{3}{2}) = \frac{2}{3}$ | M1, A1 |
$\therefore y - 0 = \frac{2}{3}(x-4)$ | M1 |
$3y = 2x - 8$ | A1 |
$2x - 3y = 8$ | **(11)** |

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The point $A$ has coordinates $(4, 6)$.

Given that $OA$, where $O$ is the origin, is a diameter of circle $C$,

\begin{enumerate}[label=(\roman*)]
\item find an equation for $C$. [4]
\end{enumerate}

Circle $C$ crosses the $x$-axis at $O$ and at the point $B$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the coordinates of $B$. [2]
\item Find an equation for the tangent to $C$ at $B$, giving your answer in the form $ax + by = c$, where $a$, $b$ and $c$ are integers. [5]
</enumerate}

\hfill \mbox{\textit{OCR C1  Q7 [11]}}