The equation of circle which touches the lines x+2y=10,x+2y=6 and its centre lies on 2x+y=6, is
A
(3x−4)2+(3y−10)2=36
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B
(3x+4)2+(3y−10)2=365
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C
(3x−4)2+(3y−10)2=365
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D
(3x−4)2+(3y−10)2=1445
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Solution
The correct option is C(3x−4)2+(3y−10)2=365 Since circle is touching both the parallel lines, ∴2r= distance between both the lines ⇒2r=4√5⇒r=2√5
Now, the mid line between both the parallel lines will pass through the centre.
The equation of mid line is x+2y=10+62⇒x+2y=8⋯(1)
And 2x+y=6⋯(2)
Solving equation (1) and (2), we get
Centre =C(43,103)
Equation of circle is (x−43)2+(y−103)2=45⇒(3x−4)2+(3y−10)2=365