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From a point on the line 4x-3y=6 , tangents are drawn to the circle x 2 +y 2 -6x-4y+4=0 which make an angle of (tan) -1 24/7 between them. Find the co-ordinates of all such points and the equations of the tangents.

From a point on the line 4x-3y=6 , tangents are
drawn to the circle x2+y2-6x-4y+4=0 which make an angle of (tan)-1 24/7 between them. Find the co-ordinates of all such points and the equations of the tangents.

Grade:11

1 Answers

Sukant Kumar
27 Points
5 years ago
This problem looks pretty hard but can be simplified using transalation of axes to bring the circle’s center at the origin.
 
We know that the circle’s center lies at (3, 2) and its radius is 3. Hence, we shift the origin to (3, 2) and the equations transform into the following form:
x2+y2=3 – C
4x-3y=0 – l
 
Now, visualize a point on the line l extending two tangents on the circle C which makes an angle whose tangent is 24/7. Now (as in figure), angle
P1PP2 is equally divided into CPP2 and CPP1 which can be equal to \alpha.
 
Image result for external point making tangent circle
 
 \tan{2\alpha} = \frac{24}{7}
 
\frac{\tan{\alpha}}{1-\tan^2\alpha}=\frac{24}{7}
 
24\tan^2\alpha+14\tan\alpha-24=0
 
(4\tan\alpha-3)(3\tan\alpha+4)=0
 
\tan\alpha=\frac{3}{4}
(as tan\alpha must be positive, due to acuteness)
 
Now we have found the angle CPP2 which can be used to find another equation of (h, k).
 
\tan\alpha=\frac{3}{\sqrt{h^2+k^2-9}}
 
(radius=3 and length of tangent=\sqrtS1)
 
On equating the values of tan\alpha, we get another circle on which our required points lie...
 
h^2+k^2=25
 
and we already know 4h-3y=0
 
On solving, you get h=\pm3, k=\pm4
 
Hence, you can now shift back the origin and find the equations of the pairs of tangents using the standard formula pretty easily.
 
Thanks,
Shukant Pal
 
 

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