The straight line \(x = 3\) intersects the circle \[ x^2 + y^2 - 8x - 10y - 9 = 0 \] at the points \(P\) and \(Q\).
Calculate
- the coordinates of \(P\) and \(Q\),
Putting \(x = 3\), the \(y\)-values we need must satisfy the equation \[ 3^2 + y^2 - 8 \times 3 - 10y - 9 = 0. \] After rearranging this, we have that \[ y^2 - 10y - 24 = 0. \] This can be factorised as \[ (y + 2)(y - 12) = 0. \] The two solutions to this equation are \(y = 12\) and \(y = -2\) and, therefore, we have that \[ P = (3,-2) \quad\text{and}\quad Q = (3,12). \]
It does not matter which way around \(P\) and \(Q\) are.
- the equations of the tangents at \(P\) and \(Q\),…
The tangent to the circle at \(P\) is perpendicular to the radius \(OP\).
To find the gradient of the radius we need to know the co-ordinates of the centre of the circle. We can find this by putting the equation into completed the square form: \[(x-4)^2 + (y - 5)^2 = 50.\]
We have the co-ordinates of \(O (4, 5)\) and of \(P (3, -2)\) so the gradient of \(OP\) is \(\frac{5 - (-2)}{4 - 3} = 7\), and the gradient of the tangent is \(-\frac{1}{7}\).
Using the coordinates of P gives us an equation for the tangent of \(y + 2 = \frac{1}{7}(x - 3)\) which can be simplified to \(7y = -x - 11\).
We can do the same for \(Q\) to find that the equation of the tangent is \(7y = x+81.\)
- the coordinates of the point of intersection of the tangents.
The point of intersection \((x,y)\) must satisfy the pair of equations \[ 7y= -x - 11, 7y = x + 81. \] Thus we need to solve \[ -x - 11= x + 81, \] which gives \(x = -46\), from where we find \(y = 5\) (as, with a little thought, we know it has to be).
