The tangent
As a tangent is a straight line it is described by an equation in the form \(y - b = m(x - a)\). You need both a point and the gradient to find its equation.
You are usually given the point - it's where the tangent meets the circle.
To find the gradient use the fact that the tangent is perpendicular to the radius from the point it meets the circle.
Work out the gradient of the radius (CP) at the point the tangent meets the circle. Then use the equation \({m_{CP}} \times {m_{tgt}} = - 1\) to find the gradient of the tangent.
Example
Find the equation of the tangent to the circle \({x^2} + {y^2} - 2x - 2y - 23 = 0\) at the point \(P(5, - 2)\) which lies on the circle.
The centre of the circle is \((1,1)\)
\({m_{CP}} = \frac{{ - 2 - 1}}{{5 - 1}} = - \frac{3}{4}\)
Hence \({m_{tgt}} = \frac{4}{3}\) since \({m_{CP}} \times {m_{tgt}} = - 1\)
So the equation of the tangent at P is:
\(y - ( - 2) = \frac{4}{3}(x - 5)\)
\(3(y + 2) = 4(x - 5)\)
\(3y - 4x + 26 = 0\)
\(4x - 3y - 26 = 0\)
Question
Find the equation of the tangent to the circle \({x^2} + {y^2} - 2x - 2y - 23 = 0\) at the point \((5,4)\)
The centre of the circle is \((1,1)\)
\({m_{radius}} = \frac{{4 - 1}}{{5 - 1}} = \frac{3}{4} \Rightarrow {m_{tgt}} = - \frac{4}{3}\)
\(y - 4 = - \frac{4}{3}(x - 5)\)
\(3(y - 4) = - 4(x - 5)\)
\(3y - 12 = - 4x + 20\)
\(3y + 4x = 32\)
\(4x + 3y - 32 = 0\)
Question
Find the equation of the tangent to the circle \({x^2} + {y^2} - 2x + 5y = 0\) at the point \((2,0)\)
The centre of the circle is \(\left( {1, - \frac{5}{2}} \right)\)
\({m_{radius}} = \frac{{0 - \left( { - \frac{5}{2}} \right)}}{{2 - 1}} = \frac{5}{2} \Rightarrow {m_{tgt}} = - \frac{2}{5}\)
\(y - 0 = - \frac{2}{5}(x - 2)\)
\(5y = - 2(x - 2)\)
\(5y = - 2x + 4\)
\(5y + 2x = 4\)
\(2x + 5y - 4 = 0\)
