Write down the value of \(p\).
About this question
Question 4.1 of Mathematics Grade 12 Paper 2, NSC Examination (DBE, November 2025). It tests the Term 2 ATP topic Analytical Geometry.
- TermTerm 2
- ATP topicAnalytical Geometry
- PaperPaper 2
- Marks1 mark
\(M(p\,;\,q)\) is the centre of a circle that cuts the \(x\)-axis at A and B. C is a point such that the line from M to C is parallel to the \(y\)-axis and cuts the \(x\)-axis at D. MC cuts the circle at \(E(-6\,;\,-1)\). Tangents drawn from C touch the circle at A and B. \(AD = (q - 1)\) units.
Diagram: the circle lies to the left of the \(y\)-axis, mostly above the \(x\)-axis. A and B are on the negative \(x\)-axis with A to the left of B. MD is a vertical line from the centre down to D on chord AB, continuing to E on the circle just below the \(x\)-axis and then to C below E. MA and MB are radii; CA and CB are tangents, so AMBC forms a kite.
M lies directly above \(E(-6\,;\,-1)\) because MC is parallel to the \(y\)-axis.
\(p = -6\)
This question also appeared in
- NSC November 2024 P1, Question 5.1 (1 marks)
More Questions On Analytical Geometry
New centre \((-8\,;\,4)\), radius 5.
Closest point to the \(y\)-axis is at:
\[x = -8 + 5 = -3\]Minimum distance = 3 units
\(A(3\,;\,4)\) lies on \(y = kx + 3\):
\[4 = k(3) + 3\]\[3k = 1\]\(k = \frac{1}{3}\)
At \(B\), \(y = 0\):
\[0 = \frac{1}{3}x + 3\]\[x = -9\]\(B(-9\,;\,0)\)
Parallel line:
\[m = -\frac{4}{3}\]\[-2 = \left(-\frac{4}{3}\right)(-15) + c\]\[-2 = 20 + c\]\[c = -22\]\(y = -\frac{4}{3}x - 22\)
\(E(1\,;\,-1)\)
so \(C\hat{S}F = 90^\circ\)
\(\therefore\) CF is a diameter (chord subtends a right angle at the circle)
Centre = midpoint of CF = \(O(0\,;\,0)\) and \(r = 6\)
\(x^2 + y^2 = 36\)