Find the equation of the tangent to circle x² + y² - 6x + 4y - 12 = 0 which is parallel to line 4x +3y +5
you want a tangent with slope -4/3
the circle is
(x-3)^2 + (y+2)^2 = 25
2(x-3) + 2(y+2)y' = 0
y' = -(x-3)/(y+2)
when y' = -4/3, 3x = 4y+17
this occurs at (7,1) and (-1,-5)
The tangents are thus
4x+3y = 6±25
To find the equation of a tangent line to a circle that is parallel to a given line, we need to follow a few steps:
1. Identify the center and radius of the circle: The given equation of the circle is x² + y² - 6x + 4y - 12 = 0. We can rewrite this equation in the standard form of a circle (x - h)² + (y - k)² = r².
So, let's complete the square for x and y terms:
(x² - 6x) + (y² + 4y) = 12
(x² - 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4
(x - 3)² + (y + 2)² = 25
From this, we can see that the center of the circle is at (3, -2) and the radius is sqrt(25) = 5.
2. Find the equation of the given line in the slope-intercept form (y = mx + b):
The given line is 4x + 3y + 5 = 0. Let's rewrite this equation in slope-intercept form:
4x + 3y = -5
3y = -4x - 5
y = (-4/3)x - 5/3
The slope-intercept form gives us the slope of the line, which is m = -4/3.
3. Determine the slope of the tangent line to the circle:
A tangent to a circle is perpendicular to the radius at the point of tangency. Since we want the tangent line to be parallel to the given line, which has a slope of m = -4/3, the tangent line must also have a slope of -4/3.
4. Find the point of tangency on the circle:
To find the point of tangency, we need to find the intersection of the circle and the line perpendicular to the given line and passing through the center of the circle.
The slope of the line perpendicular to the given line is the negative reciprocal of the slope of that line, which is m = 3/4.
Since this line passes through the center of the circle at (3, -2), we can use the point-slope form of a line to find the equation of the line:
y - (-2) = (3/4)(x - 3)
y + 2 = (3/4)x - (9/4)
y = (3/4)x - (17/4)
To find the point of tangency, we need to solve the system of equations between the circle and the perpendicular line:
(x - 3)² + (y + 2)² = 25
y = (3/4)x - (17/4)
By substituting y in terms of x in the circle equation, we can solve for x:
(x - 3)² + ((3/4)x - (17/4) + 2)² = 25
Now, we solve this equation to find the value(s) of x. Once we have the x-coordinate(s), we can substitute them back into the line equation y = (3/4)x - (17/4) to get the y-coordinate(s).
5. Use the point of tangency and the slope of the tangent line to find the equation:
Once we have the point of tangency, we can apply the point-slope form of the equation of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
where (x1, y1) is the point of tangency and m is the slope of the tangent line.
By plugging in the values of x1, y1, and m into the equation, we can determine the equation of the tangent line.
Following these steps, you should be able to find the equation of the tangent to the given circle that is parallel to the given line.