The equation of circle which touches the straight lines 2x−3y=10,2x−3y=4 and its centre lies on 2x+3y=1, is
A
(x−2)2+(y+1)2=3613
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B
(x−2)2+(y+1)2=913
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C
(x−3)2+(y+2)2=913
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D
(x−3)2+(y+2)2=3613
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Solution
The correct option is B(x−2)2+(y+1)2=913 Given lines are 2x−3y=10 and 2x−3y=4 Both touching lines are parallel. ∴2r= distance between parallel lines ⇒2r=6√13⇒r=3√13
The mid line between both parallel lines will pass through the centre and the equation of the line is 2x−3y=10+422x−3y=7⋯(1)
Solving (1) with 2x+3y=1, we get centre of the circle, C=(2,−1)