Write down the size of \(C\hat{E}M\).
About this question
Question 4.1 of Mathematics Grade 12 Paper 2, NSC Examination (May/June) (DBE, May/June 2025). It tests the Term 2 ATP topic Analytical Geometry.
- TermTerm 2
- ATP topicAnalytical Geometry
- PaperPaper 2
- Marks1 mark
M is the centre of the circle with equation \((x - 3)^2 + y^2 = 25\). \(E(-1\,;\,3)\) and T are points on the circle. EC is a tangent to the circle at E and cuts the \(x\)-axis at D. \(ED = \frac{15}{4}\) units. MT is produced to meet the tangent at \(C(-7\,;\,p)\).
Diagram: M lies on the positive \(x\)-axis. E is on the circle in the second quadrant, just left of the \(y\)-axis. The tangent at E slopes upward to the right and passes through D on the negative \(x\)-axis and \(C(-7\,;\,p)\) in the third quadrant. The line from M through T (a point on the lower left of the circle) goes on to C. Radius ME is drawn.
The tangent is perpendicular to the radius at the point of contact.
\(C\hat{E}M = 90^\circ\)
More Questions On Analytical Geometry
\(\therefore PT \perp RT\)
\(S(-12\,;\,1)\) and \(T(0\,;\,10)\)
\[m_{ST} = \frac{10 - 1}{0 - (-12)} = \frac{9}{12} = \frac{3}{4}\]\(VR \perp ST\), so
\[m_{VR} = -\frac{4}{3}\]Substitute \(R(-5\,;\,0)\):
\[0 = -\frac{4}{3}(-5) + c\]\[c = -\frac{20}{3}\]\(y = -\frac{4}{3}x - \frac{20}{3}\)
At D, \(y = 0\):
\[0 = \frac{4}{3}x + \frac{13}{3}\]\[x = -\frac{13}{4}\]\[DM = 3 - \left(-\frac{13}{4}\right) = \frac{25}{4}\]\(DM = \frac{25}{4} = 6{,}25\) units
Centre N = midpoint of AC:
\[N = \left(\frac{-3 + 2}{2}\,;\,\frac{-1 + 3}{2}\right) = \left(-\frac{1}{2}\,;\,1\right)\]\[AC = \sqrt{(2 - (-3))^2 + (3 - (-1))^2} = \sqrt{41}\]so \(r = \frac{\sqrt{41}}{2}\) and \(r^2 = \frac{41}{4}\)
\[\left(x + \frac{1}{2}\right)^2 + (y - 1)^2 = \frac{41}{4}\]\[x^2 + x + \frac{1}{4} + y^2 - 2y + 1 = \frac{41}{4}\]\[x^2 + x + y^2 - 2y - 9 = 0\]At the intersection points both equations hold, so equate:
\[x^2 + x + y^2 - 2y - 9 = x^2 - 4x + y^2 - 8y - 5\]\[5x + 6y - 4 = 0\]\(6y + 5x - 4 = 0\), i.e. \(y = -\frac{5}{6}x + \frac{2}{3}\)
\(m_{TR} = \frac{3}{4}\)
\(y = -\frac{3}{4}x - \frac{27}{4}\)