If 3x + y + k = 0 is a tangent to the circle | vedclass

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(C) The equation of the line is $3x + y + k = 0$. The equation of the circle is $x^{2} + y^{2} = 10$,which has its center at $(0, 0)$ and radius $r =

Vedclass · Accepted Answer

$\pm 10$

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(C) The equation of the line is $3x + y + k = 0$. The equation of the circle is $x^{2} + y^{2} = 10$,which has its center at $(0, 0)$ and radius $r = \sqrt{10}$. If a line $Ax + By + C = 0$ is tangent to a circle with center $(x_{1}, y_{1})$ and radius $r$,then the perpendicular distance from the center to the line must equal the radius: $\left| \frac{Ax_{1} + By_{1} + C}{\sqrt{A^{2} + B^{2}}} ight| = r$ Substituting the values: $\left| \frac{3(0) + 1(0) + k}{\sqrt{3^{2} + 1^{2}}} ight| = \sqrt{10}$ $\left| \frac{k}{\sqrt{9 + 1}} ight| = \sqrt{10}$ $\left| \frac{k}{\sqrt{10}} ight| = \sqrt{10}$ $|k| = \sqrt{10} 	imes \sqrt{10}$ $|k| = 10$ Therefore,$k = \pm 10$.

If $3x + y + k = 0$ is a tangent to the circle $x^{2} + y^{2} = 10$,the values of $k$ are

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