Let S1 : x2 + y2 = 9 and S2 : (x - 2)2 + y2 = 1.

Question

Let S1 : x2 + y2 = 9 and S2 : (x - 2)2 + y2 = 1. Then the locus of center of a variable circle S which touches S1 internally and S2 externally always passes through the points :

Accepted Answer

S1 : x2 + y2 = 9 ; C1 (0, 0), r1 = 3

S2 : (x - 2)2 + y2 = 1 ; C2 (2, 0), r2 = 1

Image

Let the variable circle S and its radius is r units.

Here S and S1 touches internally

$$ \therefore $$ Distance between center,

S + S1 = PC1 = 3 - r

Here S and S2 touches externally

$$ \therefore $$ Distance between center,

S + S2 = PC2 = 1 + r

$$ \therefore $$ PC1 + PC2 = 4 > C1 C2

So locus is ellipse whose focii are C1 \& C2 and major axis is 2a = 4 and 2ae = C1C2 = 2

$$ \Rightarrow $$ $$e = {1 \over 2}$$

$$ \Rightarrow $$ $${b^2} = 4\left( {1 - {1 \over 4}} \right) = 3$$

Centre of ellipse is midpoint of C1 \& C2 is (1, 0)

Equation of ellipse is $${{{{(x - 1)}^2}} \over {{2^2}}} + {{{y^2}} \over {{{\left( {\sqrt 3 } \right)}^2}}} = 1$$

Now by cross checking the option $$\left( {2, \pm {3 \over 2}} \right)$$ satisfied it.

Solutions · 1