Question

Given a circle with equation `x^2+(y-1)^2 = 10`, what are the equations of the two tangents to the circle that run through the point `(4, -1)` ?

Answer 1

`y = -3x+11" "` and `" "y = 1/3x-7/3`

Explanation:

This question is not quite as mean as it appears at first glance.

The points where the tangents touch the circle are `(3, 2)` and `(1, -2)`, forming a square with the centre of the circle `(0, 1)` and the specified point `(4, -1)`.

The geometry largely concerns right angled triangles with sides `1`, `3` and `sqrt(10)`.

For example, the radial line segment from `(0, 1)` to `(3, 2)` is perpendicular to the line segment between `(3, 2)` and `(4, -1)`, so this latter line segment is part of a tangent to the circle.

The line through `(3, 2)` and `(4, -1)` has equation:

`y = -3x+11`

The line through `(1, -2)` and `(4, -1)` has equation:

`y = 1/3x-7/3`

In both of these cases, calculate the slope (the coefficient of the `x` term) first, then find the intercept value to pass through one of the points.

graph{(x^2+(y-1)^2-10)(x^2+(y-1)^2-0.02)((x-4)^2+(y+1)^2-0.02)(y+3x-11)(y-1/3x+7/3)((x-3)^2+(y-2)^2-0.02)((x-1)^2+(y+2)^2-0.02) = 0 [-10, 10, -5, 5]}