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Question

The equation of circle which touches the straight lines 2x3y=10,2x3y=4 and its centre lies on 2x+3y=1, is

A
(x2)2+(y+1)2=3613
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B
(x2)2+(y+1)2=913
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C
(x3)2+(y+2)2=913
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D
(x3)2+(y+2)2=3613
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Solution

The correct option is B (x2)2+(y+1)2=913
Given lines are 2x3y=10 and 2x3y=4
Both touching lines are parallel.
2r= distance between parallel lines
2r=613r=313

The mid line between both parallel lines will pass through the centre and the equation of the line is
2x3y=10+422x3y=7 (1)

Solving (1) with 2x+3y=1, we get centre of the circle, C=(2,1)

Equation of the circle is (x2)2+(y+1)2=913

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