All Conix!
Let C 1 and C 2 be two circles. The locus of centre of the circle C which touches C 1 and C 2 externally is a
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3 solutions
The hyperbola is the only one that is a curve but has straight line asymptotes.
So? Can you explain where you apply the asymptote concept?
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As the radius of the locus circle becomes very large, it approaches becoming a line tangent to the other two circles, and the locus would be normal to that line with a fixed slope. Hence, straight line asymptotes. This is not proof that the locus is a hyperbola, it's simply enough to know which of the multiple choices is correct.
You and others here have already offered nice proofs that it's a hyperbola.
The difference in distances of the center of C from the centers (focii) of C 1 a n d C 2 is constant. If O , O 1 , a n d O 2 a r e t h e i r r e s p e c t i c e c e n t e r s ∣ O O 1 − O O 2 ∣ = c o n s t a n t = ∣ r 1 − r 2 ∣ ∣ r 1 − r 2 ∣ < ∣ r 1 + r 2 ∣ . So the locus is a hyperbola with O 1 a n d O 2 a s f o c i i .
I used this same method!
You should say
∣ O C 1 − O C 2 ∣ = c o n s t a n t And, ∣ O C 1 − O C 2 ∣ < O 1 O 2
And thus a Hyperbola.
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Thanks. I have corrected.
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You can't say O 1 O 2 = ∣ r 1 + r 2 ∣ !
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@Kishore S. Shenoy – I did not understand what you mean. I think I have not said O 1 O 2 = ∣ r 1 + r 2 ∣ . But what is wrong in it? Please explain. Thanks.
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@Niranjan Khanderia – O1O2 not equal to r1 +r2
We can get that
∣ O O 1 − O O 2 ∣ = ∣ ( r 1 + r ) − ( r 2 + r ) ∣ = ∣ r 1 − r 2 ∣ = constant .
And
∣ O O 1 − O O 2 ∣ < O 1 O 2 ( ∵ O 1 O 2 ≥ r 1 + r 2 ) ,
Thus making it a Hyperbola