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Question

If the circles x2+y24x6y12=0 and 5(x2+y2)8x14y32=0 touch each other then their point of contact is

A
(0,0)
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B
(1,1)
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C
(1,1)
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D
(0,1)
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Solution

The correct option is C (1,1)
Equations of the circles are x2+y24x6y12=0 and x2+y285x145y325=0
Centres are A(2,3) , B(45,75)

r1=4+9+12=5
r2=1625+4925+325=3

Point of contact is E.C.S (External Centre of Similitude), which divides ¯¯¯¯¯¯¯¯AB in the ratio r1:r2 externally.

So, point of contact is (462,792)=(1,1)


Alternate Solution:
Equations of the circles are x2+y24x6y12=0 and x2+y285x145y325=0
Centres are A(2,3) , B(45,75)
r1=4+9+12=5

r2=1625+4925+325=3

AB=(245)2+(375)2 =3625+6425=10025=2

AB=r1r2 the circles touch internally.
Equation of common tangent of two circle touching each other, is
S2S1=0
12x5+16y5+285=0
3x+4y+7=0 (i)
Equation of line passing through their centres
y3=43(x2)
3y4x1=0 (ii)
From equation (i) and (ii)
x=1 and y=1
So, the point of contact is (1,1)


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