Tangents are drawn from an external point #(x_1,y_1)# to the circle #x^2+y^2=a^2#. How do we find the angle between the two tangents?

1 Answer
CW
May 10, 2017

#2*sin^-1(a/sqrt(x_1^2+y_1^2))#

Explanation:

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Given equation of the circle : #x^2+y^2=a^2#,
#=> center =(0,0), radius =a#
Let #O# be the center of the circle, and #theta# be the angles between the two tangents #(angleAPB)#, as shown in the figure.
As #anglePAO=anglePBO=90^@, and AO=BO=a#
#=> DeltaPAO and DeltaPBO# are congruent.
#=> PA=PB, and PO# bisects #angleAPB#
#=> angleAPO=angleBPO=theta/2#

As #PO=sqrt(x_1^2+y_1^2)#
#=> sin(theta/2)=a/(PO)=a/sqrt(x_1^2+y_1^2)#

#=> theta/2=sin^-1(a/sqrt(x_1^2+y_1^2))#

#=> theta = 2* sin^-1(a/sqrt(x_1^2+y_1^2))#

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