Extension question
The line \(l\) is a tangent to the circle with centre \({C_1}\) and equation:
\({x^2} + {y^2} - 4x - 6y + 8 = 0\)
The point of contact A has coordinates \((1,5)\).
Question
a) Show that the equation of the line \(l\) is \(2y - x = 9\)
\({x^2} + {y^2} - 4x - 6y + 8 = 0\)
Centre : \({C_1}(2,3)\)
For \({C_1}(2,3)\) and \(A(1,5)\)
\({m_{CA}} = \frac{{5 - 3}}{{1 - 2}} = \frac{2}{{ - 1}} = - 2\)
\(\Rightarrow {m_1} = \frac{1}{2}\)
Point on the tangent is \(A(1,5)\)
and the gradient \(= \frac{1}{2}\)
So the equation is:
\(y - 5 = \frac{1}{2}(x - 1)\)
\(2y - 10 = x - 1\)
\(2y - x = 9\)
The circle with centre \({C_2}\) has equation:
\({x^2} + {y^2} + 2x + 2y - 18 = 0\)
Question
b) Show that the line \(l\) is also a tangent to this circle.
For intersection of line and circle solve:
\(2y - x = 9\)
\({x^2} + {y^2} + 2x + 2y - 18 = 0\)
Substitute \(x = 2y - 9\) in the circle equation.
\({(2y - 9)^2} + {y^2} + 2(2y - 9) + 2y - 18 = 0\)
\(4{y^2}-36y+81+{y^2}+4y-18+2y-18=0\)
\(5{y^2} - 30y + 45 = 0\)
\(5({y^2} - 6y + 9) = 0\)
\(5(y - 3)(y - 3) = 0\)
\(y = 3\)
And since there is only one solution, the line is a tangent to the circle.
You could have also shown this using the discriminant.
Question
c) If B is the point of contact, find the exact length of AB.
When \(y = 3\)
\(x = 2 \times 3 - 9 = - 3\)
so the point of contact is \(B( - 3,3)\)
For \(A(1,5)\) and \(B( - 3,3)\)
\(AB = \sqrt {1 - {{( - 3)}^2} + {{(5 - 3)}^2}}\)
\(= \sqrt {{4^2} + {2^2}} = \sqrt {20} = 2\sqrt 5\)
