If the straight line y = mx + c touches the circle x2+y2−4y=0, then the value of c will be
A
1+√1+m2
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B
1−√m2+1
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C
2(1+√1+m2)
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D
2+√1+m2
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Solution
The correct option is C2(1+√1+m2) Apply for tangency of line, centre being (0, 2) and radius = 2 ∣∣∣−2+c√1+m2∣∣∣=2⇒c2−4c+4=4+4m2 ⇒c=4±√16+16m22orc=2±2√1+m2
Most correct answer is c=2(1+√1+m2).