circle O has a diameter D and circle P has a diameter d. the two circles are externally tangent. find the length of the common external tangent in terms of the circles' diameters
let D/2 = x, and d/2 = y , where D > d
make a sketch, draw in the common tangent, join OP, and draw a line parallel to the tangent from P.
I see a rectangle and a right-angled triangle. The hypotenuse H of the right-angled triangle is equal to the length of the common tangent.
H^2 = (x+y)^2 + (x-y)^2
= x^2 + 2xy + y^2 + x^2 - 2xy + y^2 = 2x^2 + 2y^2
= 2(D^2/4) + 2(d^2/4)
= (1/2)(d^2 + D^2)
H = common tangent = (1/√2)√(D^2 + d^2)
To find the length of the common external tangent, we can use the concept of similar triangles.
Let's consider the right triangle formed by the centers of the two circles (O and P), and the point where the common external tangent touches circle O (let's call it A).
Since OA is the radius of circle O, it has a length of D/2.
Similarly, let's consider the right triangle formed by the centers of the two circles (O and P), and the point where the common external tangent touches circle P (let's call it B).
Since OB is the radius of circle P, it has a length of d/2.
Now, the line passing through the centers of two tangent circles is perpendicular to the common external tangent.
By connecting the centers of the two circles (O and P) and the point of tangency on both circles, we form a right triangle (let's call it triangle OAP).
By applying the Pythagorean theorem, we have:
(OA)^2 + (AP)^2 = (OP)^2
Since OA is D/2 and AP is the common external tangent length that we need to find, we can rewrite the equation as:
(D/2)^2 + (AP)^2 = (D/2 + d/2)^2
Simplifying the equation further:
(D^2/2^2) + (AP)^2 = [(D + d)/2]^2
(D^2/4) + (AP)^2 = (D^2 + 2Dd + d^2)/4
(AP)^2 = (D^2 + 2Dd + d^2)/4 - (D^2/4)
(AP)^2 = (2Dd + d^2)/4
Simplifying and taking the square root of both sides:
AP = √[(2Dd + d^2)/4]
AP = √[(Dd + d^2/2)]
Therefore, the length of the common external tangent, AP, between the two circles is √[(Dd + d^2/2)].