If the line √2x + k = 0 touches the circle x² + y² = 9, then find the value of k.

Steps to Solve

1. Identify the center and radius of the circle
   The equation of the circle is given by (x^2 + y^2 = 9).
   Here, the center is ((0, 0)) and the radius (r) is given by:
   $$ r = \sqrt{9} = 3 $$

2. Rewrite the line equation in slope-intercept form
   The line is given by (\sqrt{2}x + k = 0).
   Rearranging this gives:
   $$ y = -\frac{\sqrt{2}}{1}x - k $$
   This shows that the line has a slope of (-\sqrt{2}).

3. Find the perpendicular distance from the center of the circle to the line
   The formula for the distance (d) from a point ((x_0, y_0)) to a line (Ax + By + C = 0) is given by:
   $$ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} $$
   Here, (A = \sqrt{2}), (B = 0), (C = k), and the center of the circle is ((0, 0)).
   Thus, substituting these values, we get:
   $$ d = \frac{|\sqrt{2}(0) + 0(0) + k|}{\sqrt{(\sqrt{2})^2 + 0^2}} = \frac{|k|}{\sqrt{2}} $$

4. Set the distance equal to the radius for tangency
   For the line to touch the circle (tangency condition), the distance must equal the radius:
   $$ \frac{|k|}{\sqrt{2}} = 3 $$

5. Solve for (k)
   Multiplying both sides by (\sqrt{2}):
   $$ |k| = 3\sqrt{2} $$
   Thus, (k) can be either positive or negative:
   $$ k = 3\sqrt{2} \quad \text{or} \quad k = -3\sqrt{2} $$