The common external tangents to the graphs for (x+1)^2+y^2=1 and (x-2)^2+y^2=4 are drawn. Find the slope of the tangent line that has a positive slope.
I tried finding the derivatives and setting them equal to each other, but that didn't give me an answer. Please help!
I agree that the calculus gets messy, but the algebra is simple.
Label the centers C and D, respectively, and the points of tangency E and F.
We want the slope of the line EF.
Draw CE and DF. Since they are both perpendicular to EF, they are parallel.
Mark point P on DF such that CE is parallel to EF. So, CD = PF = 1, meaning that PD=1 as well.
CD = 1+2 = 3, so our line makes an angle z with the x-axis, such that
sin(z) = 1/3, so our slope, tan(z) is 1/√8.
To find the common external tangents to the given graphs, we can start by finding their equations.
The first equation, (x+1)^2+y^2=1, represents a circle centered at (-1, 0) with a radius of 1.
The second equation, (x-2)^2+y^2=4, represents a circle centered at (2, 0) with a radius of 2.
To find the external tangents, we need to find the lines that are tangent to both circles. There are two common external tangents, one on each side of the circles.
To find the slope of these tangents, we can use the fact that the tangent line is perpendicular to the radius of the circle at the point of tangency.
Let's start by finding the slope for the tangent line to the first circle at the point of tangency. We can find the derivative of the equation (x+1)^2+y^2=1 and use it to find the slope of the tangent line.
Differentiating the equation implicitly, we get:
2(x+1)dx + 2ydy = 0
Rearranging and solving for dy/dx, we have:
dy/dx = -(x+1)/y
Now we can find the slope of the tangent line at the point of tangency for the first circle by substituting the x-coordinate of the point of tangency (-1) into the derivative:
dy/dx = -(-1+1)/y
dy/dx = 0/y
dy/dx = 0
The slope of the tangent line for the first circle is 0.
Similarly, we can find the slope of the tangent line for the second circle at the point of tangency. Differentiating the equation (x-2)^2+y^2=4, we have:
2(x-2)dx + 2ydy = 0
Rearranging and solving for dy/dx, we get:
dy/dx = -(x-2)/y
Now we can find the slope of the tangent line at the point of tangency for the second circle by substituting the x-coordinate of the point of tangency (2) into the derivative:
dy/dx = -(2-2)/y
dy/dx = 0/y
dy/dx = 0
The slope of the tangent line for the second circle is also 0.
Since both the tangent lines have a slope of 0, this means they are both horizontal lines. Therefore, there are no common external tangents with a positive slope for the given circles.
Hence, there is no tangent line with a positive slope that is common to both circles.