Asked by Anon
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 tried finding the derivatives and setting them equal to each other, but that didn't give me an answer. Please help!
Answers
Answered by
Steve
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.
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.
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