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)# ?

1 Answer
May 22, 2015

Answer:

#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]}