At the points of intersection of circle #x^2+y^2-y-3=0# and #x=1#, two tangents are drawn. What is the point of intersection of the tangents and angle between them?

2 Answers
Shwetank Mauria
Jun 24, 2017

Angle between tangents is #67.38^@#

Explanation:

As the circle is #x^2+y^2-y-3=0#, the centre is #(0,1/2)# and when #x=1#, we have

#y^2-y-2=0# i.e. #(y-2)(y+1)=0# i.e. #y=2# and #-1#

Hence we are seeking tangent at #(1,-1)# and #(1,2)#

The slope of radius joining centre #(0,1/2)# and #(1,-1)# is #(-1-1/2)/(1-0)=-3/2# and hence slope of tangent at #(1,-1)# would be #2/3#. Hence tangent is #y+1=2/3(x-1)# or #3y+3=2x-2# i.e. #2x-3y-5=0#.

Similarly, slope of radius joining centre #(0,1/2)# and #(1,2)# is #(2-1/2)/(1-0)=3/2# and hence slope of tangent at #(1,2)# would be #-2/3#. Hence tangent is #y-2=-2/3(x-1)# or #3y-6=-2x+2# i.e. #2x+3y-8=0#.

Now the slope of one tangent is #m_1=2/3# and that of other is #m_2=-2/3# and hence tangent of angle between them is

#(m_1-m_2)/(1+m_1m_2)=(2/3-(-2/3))/(1+2/3xx(-2/3))=(4/3)/(1-4/9)=4/3xx9/5=12/5=2.4#

and angle is #tan^(-1)2.4=67.38^@#

graph{(x^2+y^2-y-3)(2x-3y-5)(2x+3y-8)=0 [-5, 5, -2, 3]}

Narad T.
Jun 24, 2017

The point of intersection is #(3.25,0.5)# and the angle is #=67.4º#

Explanation:

Let`s rewrite the equation of the circle

#x^2+y^2-y-3=0#

We complete the squares and factorise

#x^2+y^2-y+1/4=3+1/4#

#x^2+(y-1/2)^2=13/4#

The center of the circle is #(0,1/2)# and the radius is #r=sqrt13/2#

Let #x=1#

#1+(y-1/2)^2=13/4#

#(y-1/2)^2=13/4-1=9/4#

#y-1/2=+-3/2#

#y=2# or #y=-1#

The coordinates of the tangents are #A=(1,2)# and #A'=(1,-1)#

The slopes of the radius from the center to #A# and #A'# are

#=(2-0.5)/(1-0)=3/2#

and

#=(0.5-(-1))/(0-1)=-3/2#

The slopes of the tangents are

#m_1=-2/3#

and

#m_2=2/3#

The equations of the tangents are

#y-2=-2/3(x-1)#

#=>#, #3y-6=-2x+2#,

#=>#, #3y+2x=8#...........(1)

and

#y+1=2/3(x-1)#

#=>#, #3y+3=2x-2#

#=>#, #3y-2x=-5#............(2)

Solving for #(x,y)# in equations #(1)# and #(2)#, we get

#6y=3#, #=>#, #y=3/6=0.5#

#2x=8-3y=8-1.5=6.5#

#x=6.5/2=3.25#

The point of intersection is #(3.25,0.5)#

The angle between the tangents is

#tantheta=|(m_1-m_2)/(1+m_1m_2)|#

#=|(-2/3-2/3)/(1-2/3*2/3)|#

#=|(-4/3)/(5/9)|#

#=12/5#

Therefore,

#theta=67.4º#
graph{(x^2+y^2-y-3)(y+1-2/3(x-1))(y-2+2/3(x-1))(y-0.5)=0 [-6.24, 6.243, -3.12, 3.12]}

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