Find the equations of tangents from point #(-2,11)# to the circle #x^2+y^2=25#?

2 Answers
Jan 11, 2018

Please see below.

Explanation:

A quick sketch (or even a mental image) will show us that
The point #(-2,11)# is above the circle in the second quadrant.
We expect two solutions, one tangent to the circle in the second quadrant and one in the first.
Clearly then, we need only consider the upper semicircle whose equation is:

#y = sqrt(25-x^2)#.

So #dy/dx = -x/sqrt(25-x^2)#. #" "# (Use the power and chain rules.)

The point of tangency is a point on the semicircle. If we call the #x#-coordinate #a#, then the #y#-coordinate is #sqrt(25-a^2)#.
The slope of the tangent at that point is #m = -a/sqrt(25-a^2)#.

The equation of a line tangent to the semicircle at the point where #x=a# is

#y - sqrt(25-a^2) = -a/sqrt(25-a^2)(x-a)#.

We want #(-2,11)# to lie on this line so we want

#11 - sqrt(25-a^2) = -a/sqrt(25-a^2)(-2-a)#.

Get rid of the fraction by multiplying by #sqrt(25-a^2)#. (I also factored the minus sign out of #(-2-a)#.)

#11sqrt(25-a^2) - (25-a^2) = a(a+2)# #" "# Isolate the square root term.

#11sqrt(25-a^2) = 2a+25#.

Square both sides and solve the resulting quadratic to get

#a = -24/5# #" "# or #" "# #a=4#

Recall that #y=sqrt(25-x^2)#, so the points of tangency are

#(-24/5, 7/5)# and #(4,3)#.

The tangent line through #(-24/5, 7/5)# has slope #m=24/7#.

And the tangent line through #(4,3)# has slope #m=-4/3#.

Writing the equation of these two lines is left to the reader.

The points of tangency are not clear in this graph, but you can zoom in and out and drag the graph around in the window.
graph{(x^2+y^2-25)(y-11-24/7(x+2))(y-11+4/3(x+2))=0 [-22.47, 28.9, -9.22, 16.44]}

This picture (using Desmos) may be more clear, but you cannot change the view.
enter image source here

Jan 11, 2018

The tangents are #4x+3y-25=0# and #24x-7y+125=0#

Explanation:

Let the slope of tangent be #m#. As the tangent passes through the point #(-2,11)#, its equation is

#y-11=m(x+2)# or #y=mx+2m+11#

Now this cuts the circle #x^2+y^2=25# at points given by

#x^2+(mx+2m+11)^2=25#

or #x^2+m^2x^2+4m^2+121+4m^2x+22mx+44m-25=0#

or #x^2(1+m^2)+x(4m^2+22m)+4m^2+44m+96=0#

The above quadratic equation can have up to two solutions. We have two real roots when discriminant is positive and it shows the line cutting the circle at two distinct points. When discriminant is negative, we do not have any real root, which represents the line not touching the circle at all.

And when the line touches the circle, there is only one root (or two coincident roots) of quadratic equation, which happens when the discriminant is equal to zero. As the line we are seeking is a tangent i.e. touches the circle, the discriminant must be zero i.e.

#(4m^2+22m)^2-4(1+m^2)(4m^2+44m+96)=0#

or #color(blue)(16m^4)+484m^2color(blue)(+176m^3)color(red)(-16m^4-176m^3)-384m^2-16m^2-176m-384=0#

i.e. #84m^2-176m-384=0# or #21m^2-44m-96=0#

or #21m^2+28m-72m-96=0#

or #(7m-24)(3m+4)=0#

Hence #m=-4/3# or #24/7#

and tangents are #y=-4/3x-8/3+11# i.e. #4x+3y-25=0#

and #y=24/7x+48/7+11# or #24x-7y+125=0#

graph{(x^2+y^2-25)(24x-7y+125)(4x+3y-25)=0 [-19.92, 20.08, -7.6, 12.4]}