Two lines emerging from Point D (-4,-4) are inclined at 42 °to each other....... ?
Two lines emerging from Point D (-4,-4) are inclined at 42 °to each other. Bisector of this angle intersects the x-axis at P (-4,0). Two circles touch each other externally at P. Calculate area trapped between the two circles and the two tangents.
Update:EDIT: The two circles touch the two lines too (refer to the figure below)
Update 3:az_lender : Plz click to the following linc:
https://www.quora.com/What-is-the-area-trapped-bet...
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Answers & Comments
First let's move the y-axis over to coincide with the line DP. The x = -4 is just an added complication that makes the problem more confusing. With DP corrected to be x=0, the equation of the positively sloped diagonal becomes y = x*tan(90 - 21) - 4 = x*tan(69 deg) - 4.
Next I seek the equation of the upper circle. We know its equation is of the form x^2 + (y-k)^2 = k^2, and that dy/dx = tan(69 deg) where the circle intersects the positively sloped diagonal. On the circle, you have
2x + 2(y-k)*(dy/dx) = 0 =>
dy/dx = x/(k - y), but you also know that this is tan(69 deg), so you have:
x/(k - y) = tan(69) and y = x*tan(69) - 4.
The equation of the circle itself simplifies to
x^2 = (2k - y)*y, so you have 3 equations in 3 unknowns, and with some ugliness should get a solution for k.
Similarly, you can find the coordinates of the center and intersection for the lower circle; and finally, integrate to get the yellow area.