Additional Mathematics – Coordinate geometry of the circle | e-Consult

Rewrite the circle in centre‑radius form.

\[

x^{2}-4x+y^{2}-6y+9=0\;\Longrightarrow\;(x-2)^{2}+(y-3)^{2}=4.

\]

Thus centre \(C(2,3)\) and radius \(r=2\).

Let the equation of a line through \(P(1,5)\) be \(y-5=m(x-1)\). The perpendicular distance from \(C\) to this line must equal \(r\):

\[

\frac{|m(2-1)-(3-5)|}{\sqrt{m^{2}+1}}=2

\;\Longrightarrow\;

\frac{|m+2|}{\sqrt{m^{2}+1}}=2.

\]

Square:

\[

\frac{(m+2)^{2}}{m^{2}+1}=4\;\Longrightarrow\;(m+2)^{2}=4(m^{2}+1)

\]

\[

m^{2}+4m+4=4m^{2}+4\;\Longrightarrow\;3m^{2}-4m=0\;\Longrightarrow\;m(3m-4)=0.

\]

Hence \(m=0\) or \(m=\dfrac{4}{3}\).

Corresponding lines through \(P\):

• For \(m=0\): \(y-5=0\Rightarrow y=5\).
• For \(m=\dfrac{4}{3}\): \(y-5=\dfrac{4}{3}(x-1)\Rightarrow 3y-15=4x-4\Rightarrow 4x-3y+11=0.\)

Find points of tangency by solving each line with the circle.

1. Line \(y=5\):
   

   Substitute into \((x-2)^{2}+(5-3)^{2}=4\):

   \((x-2)^{2}+4=4\Rightarrow (x-2)^{2}=0\Rightarrow x=2.\)
   

   Point of tangency \(T_{1}(2,5).\)

2. Line \(4x-3y+11=0\):
   

   Write \(y=\frac{4x+11}{3}\) and substitute:
   

   \((x-2)^{2}+\Bigl(\frac{4x+11}{3}-3\Bigr)^{2}=4.\)
   

   Simplify \(\frac{4x+11}{3}-3=\frac{4x+11-9}{3}=\frac{4x+2}{3}\).
   

   Equation becomes \((x-2)^{2}+\Bigl(\frac{4x+2}{3}\Bigr)^{2}=4.\)
   

   Multiply by 9: \(9(x-2)^{2}+(4x+2)^{2}=36.\)
   

   Expand: \(9(x^{2}-4x+4)+(16x^{2}+16x+4)=36\)
   

   \(9x^{2}-36x+36+16x^{2}+16x+4=36\)
   

   \(25x^{2}-20x+40=36\Rightarrow 25x^{2}-20x+4=0.\)
   

   Divide by 1: \(25x^{2}-20x+4=0\). Discriminant \(D= (-20)^{2}-4\cdot25\cdot4=400-400=0\), confirming tangency.
   

   Thus \(x=\frac{20}{2\cdot25}= \frac{20}{50}=0.4.\)
   

   Then \(y=\frac{4(0.4)+11}{3}= \frac{1.6+11}{3}= \frac{12.6}{3}=4.2.\)
   

   Point of tangency \(T_{2}(0.4,4.2).\)

Hence the two tangents are

• \(\boxed{y=5}\) touching at \((2,5)\).
• \(\boxed{4x-3y+11=0}\) touching at \((0.4,\;4.2).\)