SPS SPS SM 2020 October — Question 8 10 marks

Exam BoardSPS
ModuleSPS SM (SPS SM)
Year2020
SessionOctober
Marks10
TopicCircles
TypeFind centre and radius from equation
DifficultyModerate -0.8 This is a straightforward multi-part circle question requiring completing the square (standard technique), solving simultaneous equations by substitution, and using the midpoint property of diameters. All parts are routine textbook exercises with no problem-solving insight required, making it easier than average but not trivial due to the algebraic manipulation involved.
Spec1.02c Simultaneous equations: two variables by elimination and substitution1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle

The equation of a circle is \(x^2 + y^2 + 6x - 2y - 10 = 0\).
  1. Find the centre and radius of the circle. [3]
  2. Find the coordinates of any points where the line \(y = 2x - 3\) meets the circle \(x^2 + y^2 + 6x - 2y - 10 = 0\). [4]
  3. State what can be deduced from the answer to part ii. about the line \(y = 2x - 3\) and the circle \(x^2 + y^2 + 6x - 2y - 10 = 0\). [1]
  4. The point \(A(-1,5)\) lies on the circumference of the circle \(x^2 + y^2 + 6x - 2y - 10 = 0\). Given that \(AB\) is a diameter of the circle, find the coordinates of \(B\). [2]

The equation of a circle is $x^2 + y^2 + 6x - 2y - 10 = 0$.
\begin{enumerate}[label=(\roman*)]
\item Find the centre and radius of the circle. [3]
\item Find the coordinates of any points where the line $y = 2x - 3$ meets the circle $x^2 + y^2 + 6x - 2y - 10 = 0$. [4]
\item State what can be deduced from the answer to part ii. about the line $y = 2x - 3$ and the circle $x^2 + y^2 + 6x - 2y - 10 = 0$. [1]
\item The point $A(-1,5)$ lies on the circumference of the circle $x^2 + y^2 + 6x - 2y - 10 = 0$. Given that $AB$ is a diameter of the circle, find the coordinates of $B$. [2]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS SM 2020 Q8 [10]}}