If the line y = mx + C is a tangent to the ci | vedclass

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(B) The given equation of the circle is $x^2 + y^2 = 16$,which is of the form $x^2 + y^2 = a^2$ with $a^2 = 16$.We know that the line $y = mx + C$ is

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$\pm \frac{1}{4} \sqrt{C^2 - 16}$

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(B) The given equation of the circle is $x^2 + y^2 = 16$,which is of the form $x^2 + y^2 = a^2$ with $a^2 = 16$.We know that the line $y = mx + C$ is a tangent to the circle $x^2 + y^2 = a^2$ if and only if $C^2 = a^2(1 + m^2)$.Substituting $a^2 = 16$ into the condition,we get $C^2 = 16(1 + m^2)$.Dividing by $16$,we have $\frac{C^2}{16} = 1 + m^2$.Rearranging for $m^2$,we get $m^2 = \frac{C^2}{16} - 1 = \frac{C^2 - 16}{16}$.Taking the square root on both sides,we get $m = \pm \frac{1}{4} \sqrt{C^2 - 16}$.Thus,the correct option is $B$.

If the line $y = mx + C$ is a tangent to the circle $x^2 + y^2 = 16$,then $m =$

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