The equation of the circle which passes through (1,2) and (0,2) and has its radius as small as possible is
A
x2+y2−2x−3y−2=0
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B
x2+y2−x−4y+4=0
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C
x2+y2−3x+4=0
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D
x2+y2+x+2y+1=0
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Solution
The correct option is Bx2+y2−x−4y+4=0 The radius will be minimum, if the given points are the end points of a diameter. Equation of circle in diametric form is (x−1)(x−0)+(y−2)(y−2)=0⇒x2−x+y2−4y+4=0∴x2+y2−x−4y+4=0