Find the equations of two tangents to the circle x2 + y2 − 9 = 0 that are parallel to the line 3x + 4y = 6

To find the equations of two tangents to the circle x^2 + y^2 - 9 = 0 that are parallel to the line 3x + 4y = 6, we can follow these steps:

Step 1: Find the center and radius of the circle. The given equation of the circle is x^2 + y^2 - 9 = 0. Comparing this equation with the standard form of a circle (x - h)^2 + (y - k)^2 = r^2, we can see that the center of the circle is at (0, 0) and the radius is √9 = 3.

Step 2: Find the slope of the given line. The given line equation is 3x + 4y = 6. We can rewrite it in slope-intercept form (y = mx + b) by isolating y: 4y = -3x + 6 y = (-3/4)x + 3/2 Comparing this equation with y = mx + b, we can see that the slope of the line is -3/4.

Step 3: Find the slope of the tangent lines. Since the tangent lines are parallel to the given line, they will have the same slope of -3/4.

Step 4: Find the points of tangency. The distance between the center of the circle and the points of tangency is equal to the radius of the circle (which is 3). Let (x1, y1) and (x2, y2) be the points of tangency.

Using the point-slope form of a line, we can write the equations of the tangent lines as: y - y1 = (-3/4)(x - x1) (Equation 1) y - y2 = (-3/4)(x - x2) (Equation 2)

Step 5: Substitute the coordinates of the center into the equations. Since the center of the circle is (0, 0), we can substitute x1 = 0 and y1 = 0 into Equation 1: y - 0 = (-3/4)(x - 0) y = (-3/4)x

Similarly, we can substitute x2 = 0 and y2 = 0 into Equation 2: y - 0 = (-3/4)(x - 0) y = (-3/4)x

Therefore, the equations of the two tangents to the circle x^2 + y^2 - 9 = 0 that are parallel to the line 3x + 4y = 6 are y = (-3/4)x and y = (-3/4)x.