The straight line xcosθ+ysinθ=2, will touch the circle x2+y2−2x=0, if
A
θ=nπ,n∈Z
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B
θ=(2n+1)π,n∈Z
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C
θ=2nπ,n∈Z
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D
None of these
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Solution
The correct option is Cθ=2nπ,n∈Z x2+y2−2x=0 C=(1,0),r=√12=1
Now,
The line xcosθ+ysinθ=2 is tangent on the circle if distance from centre to line is equal to radius, so ∣∣
∣∣cosθ−2√cos2θ+sin2θ∣∣
∣∣=1⇒cosθ−2=±1⇒cosθ=1,3∴cosθ=1(∵cosθ∈[−1,1])⇒θ=2nπ,n∈Z