# 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) ?

May 22, 2015

$y = - 3 x + 11 \text{ }$ and $\text{ } y = \frac{1}{3} x - \frac{7}{3}$

#### Explanation:

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

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

The geometry largely concerns right angled triangles with sides $1$, $3$ and $\sqrt{10}$.

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

The line through $\left(3 , 2\right)$ and $\left(4 , - 1\right)$ has equation:

$y = - 3 x + 11$

The line through $\left(1 , - 2\right)$ and $\left(4 , - 1\right)$ has equation:

$y = \frac{1}{3} x - \frac{7}{3}$

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]}