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Question

The equation of circle which touches the lines x+2y=10, x+2y=6 and its centre lies on 2x+y=6, is

A
(3x4)2+(3y10)2=36
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B
(3x+4)2+(3y10)2=365
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C
(3x4)2+(3y10)2=365
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D
(3x4)2+(3y10)2=1445
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Solution

The correct option is C (3x4)2+(3y10)2=365
Since circle is touching both the parallel lines,
2r= distance between both the lines
2r=45r=25

Now, the mid line between both the parallel lines will pass through the centre.
The equation of mid line is
x+2y=10+62x+2y=8 (1)
And 2x+y=6 (2)
Solving equation (1) and (2), we get
Centre =C(43,103)

Equation of circle is
(x43)2+(y103)2=45(3x4)2+(3y10)2=365

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