16 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.You can do this by the concept of centres of similitude can you not? C is just the external centre of similitude so it divides AB in ratio 2:1 externally
i have no idea what that word means
similitude
i tried assuming two point
h k
Alpha beta
then the perp slope from center on tangets will be perp to line
there were 4 eqn and 4 variable
but damn it is near impossible to solve those equatins
C divides AB externally in the ratio rA:rB is what you need to know
I assume you got coords of B
ye i did
Now it's just applying section formula
i mean we can do it without finding tangent and by method of similar triangle easily
i just want to know
how to find tangents
if i ever have to in the future
Take general tangent equation to one circle and impose condition that it is also true for the other, I guess.
(y-yC) = m(x-xC)±r√(1+m²)
what is this? i only studied that T form for tangents
xx1+yy1 + ....
Conditions for tangency for conics?
c =r√1+m² for circle
c = a/m for parabola
c = √a²m²+b² for ellipse
c = √a²m²-b² for hyperbola ?
ooh that
It would be better to approach this by geometry imo. Don't need the equation at all.
ye ik that , similar triangles , i just want to know if in future i want to find the eqn how to do it
+solved @SirLancelotDuLac @Opt
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