If y=sqrt{3}x+k_1 and y=sqrt{3}x+k_2 are two | vedclass

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(B) The distance between two parallel lines $y=mx+k_1$ and $y=mx+k_2$ is given by $d = \frac{|k_1-k_2|}{\sqrt{1+m^2}}$.Since these lines are parallel

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) The distance between two parallel lines $y=mx+k_1$ and $y=mx+k_2$ is given by $d = \frac{|k_1-k_2|}{\sqrt{1+m^2}}$.Since these lines are parallel tangents to a circle of radius $r=2$,the distance between them must be equal to the diameter of the circle,which is $2r = 2 	imes 2 = 4$.Here,$m = \sqrt{3}$,so $m^2 = 3$.Substituting these values into the distance formula:$\frac{|k_1-k_2|}{\sqrt{1+3}} = 4$$\frac{|k_1-k_2|}{\sqrt{4}} = 4$$\frac{|k_1-k_2|}{2} = 4$$|k_1-k_2| = 8$.Thus,option $B$ is correct.

If $y=\sqrt{3}x+k_1$ and $y=\sqrt{3}x+k_2$ are two parallel tangents of a circle of radius $2 \text{ units}$,then $|k_1-k_2|$ is equal to

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