Write down the coordinates of V.
About this question
Question 4.3 of Mathematics Grade 12 Paper 2, NSC Examination (May/June) (DBE, May/June 2026). It tests the Term 2 ATP topic Analytical Geometry.
- TermTerm 2
- ATP topicAnalytical Geometry
- PaperPaper 2
- Marks2 marks for the method (formula and substitution) and the correct final answer
The circle with centre M has equation \((x - 2)^2 + (y - 4)^2 = 25\). Diameter TMR and tangent SR are drawn. The equation of line ST is \(y = 2x + 5\). ST cuts the circle again at V. VR is drawn.
Diagram: T is on the circle at the lower left, R at the upper right end of the diameter through M. Line ST rises steeply from T through V (top of the circle) to S, where it meets the tangent at R.
From 4.2 the other intersection has \(x = 2\):
\[y = 2(2) + 5 = 9\]\(V(2\,;\,9)\)
More Questions On Analytical Geometry
\(m_{TR} = \frac{3}{4}\)
S lies outside the new circle.
\(\dfrac{DB}{DC} = \dfrac{1}{3}\)
In \(\triangle ABD\), \(\alpha\) is an exterior angle:
\[\alpha = A\hat{B}D + B\hat{A}C\]\[B\hat{A}C = 63{,}43^\circ - 18{,}43^\circ\]\(B\hat{A}C = 45^\circ\)
\(C(-7\,;\,p)\) lies on EC:
\[p = \frac{4}{3}(-7) + \frac{13}{3}\]\[p = -\frac{28}{3} + \frac{13}{3} = -\frac{15}{3}\]\(p = -5\)
At \(B\), \(y = 0\):
\[0 = \frac{1}{3}x + 3\]\[x = -9\]\(B(-9\,;\,0)\)