To find the equations of two tangents to the circle x2 + y2 − 6x + 4y − 9 = 0 that are parallel to the line 3x + 4y = 6, we can follow these steps:

Step 1: Rewrite the equation of the circle in standard form by completing the square for both x and y terms:
(x^2 - 6x) + (y^2 + 4y) = 9
(x^2 - 6x + 9) + (y^2 + 4y + 4) = 9 + 9 + 4
(x - 3)^2 + (y + 2)^2 = 22

Step 2: Determine the center and radius of the circle. The center is (3, -2) and the radius is √22.

Step 3: Find the slope of the given line 3x + 4y = 6. Rearrange the equation to slope-intercept form:
4y = -3x + 6
y = (-3/4)x + 3/2

The slope of the line is -3/4.

Step 4: Since the tangents to a circle are perpendicular to the radius at the point of tangency, the slopes of the tangents will be the negative reciprocal of the slope of the radius.

The slope of the radius is the slope of the line connecting the center of the circle to the point of tangency. Since the tangents are parallel to the line 3x + 4y = 6, the slope of the radius will be -3/4.

Therefore, the slopes of the tangents will be 4/3.

Step 5: Use the point-slope form to find the equations of the tangents. We will use the point of tangency as the point on the circle.

Let (x1, y1) be the point of tangency.

Using the point-slope form, the equation of the tangent line with slope m passing through the point (x1, y1) is given by:
y - y1 = m(x - x1)

Substituting the slope m = 4/3 and the coordinates of the point of tangency (x1, y1) into the equation, we can find the equations of the two tangents.

Step 6: Solve the system of equations formed by the circle and the equations of the tangents to find the coordinates of the point of tangency.

Substitute the equation of the tangent line into the equation of the circle:
(x - 3)^2 + (y + 2)^2 = 22
(x - 3)^2 + ((4/3)(x - x1) + y1 + 2)^2 = 22

Expand and simplify the equation to solve for x:
(x - 3)^2 + (4/3)(x - x1 + 3/4)^2 = 22

Solve for x and substitute the value of x back into the equation of the tangent line to find the corresponding y-coordinate.

Step 7: Substitute the coordinates of the point of tangency into the equation of the tangent line to find the equations of the two tangents.

Repeat Step 6 for the second tangent line.

By following these steps, you will be able to find the equations of the two tangents to the given circle that are parallel to the line 3x + 4y = 6.

Question

To find the equations of two tangents to the circle x2 + y2 − 6x + 4y − 9 = 0 that are parallel to the line 3x + 4y = 6, we can follow these steps:

Step 1: Rewrite the equation of the circle in standard form by completing the square for both x and y terms:
(x^2 - 6x) + (y^2 + 4y) = 9
(x^2 - 6x + 9) + (y^2 + 4y + 4) = 9 + 9 + 4
(x - 3)^2 + (y + 2)^2 = 22

Step 2: Determine the center and radius of the circle. The center is (3, -2) and the radius is √22.

Step 3: Find the slope of the given line 3x + 4y = 6. Rearrange the equation to slope-intercept form:
4y = -3x + 6
y = (-3/4)x + 3/2

The slope of the line is -3/4.

Step 4: Since the tangents to a circle are perpendicular to the radius at the point of tangency, the slopes of the tangents will be the negative reciprocal of the slope of the radius.

The slope of the radius is the slope of the line connecting the center of the circle to the point of tangency. Since the tangents are parallel to the line 3x + 4y = 6, the slope of the radius will be -3/4.

Therefore, the slopes of the tangents will be 4/3.

Step 5: Use the point-slope form to find the equations of the tangents. We will use the point of tangency as the point on the circle.

Let (x1, y1) be the point of tangency.

Using the point-slope form, the equation of the tangent line with slope m passing through the point (x1, y1) is given by:
y - y1 = m(x - x1)

Substituting the slope m = 4/3 and the coordinates of the point of tangency (x1, y1) into the equation, we can find the equations of the two tangents.

Step 6: Solve the system of equations formed by the circle and the equations of the tangents to find the coordinates of the point of tangency.

Substitute the equation of the tangent line into the equation of the circle:
(x - 3)^2 + (y + 2)^2 = 22
(x - 3)^2 + ((4/3)(x - x1) + y1 + 2)^2 = 22

Expand and simplify the equation to solve for x:
(x - 3)^2 + (4/3)(x - x1 + 3/4)^2 = 22

Solve for x and substitute the value of x back into the equation of the tangent line to find the corresponding y-coordinate.

Step 7: Substitute the coordinates of the point of tangency into the equation of the tangent line to find the equations of the two tangents.

Repeat Step 6 for the second tangent line.

By following these steps, you will be able to find the equations of the two tangents to the given circle that are parallel to the line 3x + 4y = 6.

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