Where do this pair of tangents to a circle meet?

Where do this pair of tangents to a circle meet?

Question

The straight line $x = 3$ intersects the circle $x^2 + y^2 - 8x - 10y - 9 = 0$ at the points $P$ and $Q$.

Calculate

1. the coordinates of $P$ and $Q$,

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1. the equations of the tangents at $P$ and $Q$,

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1. the coordinates of the point of intersection of the tangents.

Solution

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.

1. 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 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).