Find the length of the tangents from the origin to the circle 3x²+3y²+12+8y+8=0. Find also the acute angle between the tangents from the origin to the circle (answer= 2√6/3 ,86°)?

1 Answer
Shwetank Mauria
Oct 5, 2016

Length of the tangents from the origin to the circle is #(2sqrt6)/3#
and acute angle between the tangents from the origin to the circle is #94.4^o#

Explanation:

The equation of the given circle cannot be #3x²+3y²+12+8y+8=0# as the radius is #sqrt(4^2-20)=sqrt(-4)#. It should rather be #3x^2+3y^2+12x+8y+8=0# or #x^2+y^2+4x+8/3y+8/3=0#

Hence the center of the circle is #(-2,-4/3)# and radius is #sqrt((-2)^2+(-4/3)^2-8/3)=1/3sqrt(36+16-24)=1/3sqrt28#.

Now let us draw tangents from origin at #(0,0)# passing through the center of given circle i.e. #(-2,-4/3)# as also join origin with center of circle at #(-2,-4/3)#. This along with radii from center of circle to points where tangents touch circle form right angle, as shown in the image shown below.
enter image source here
Now radius is #sqrt28/3# and distance between #(0,0)# and center at #(-2,-4/3)# is

#sqrt((0+2)^2+(0+4/3)^2)=sqrt(4+16/9)=sqrt(52/9)=sqrt52/3#

Hence length of tangents is #sqrt((sqrt52/3)^2-(sqrt28/3)^2)#

= #sqrt(52/9-28/9)=sqrt(24/9)=sqrt((2xx2xx6)/(3xx3))=(2sqrt6)/3#

Further let acute angle #/_P_0PC=theta#, then #sintheta=(sqrt28/3)/(sqrt52/3)=sqrt(7/13)=0.7338# and

#theta=47.2^o#

and angle between the two tangents is #47.2^o xx2=94.4^o#

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