In a triangle the sum of two sides is x and the product of the same two sides is y. If x² - c² = y, where c is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is?

We know, in-radius=area/semi-perimeter=Δ/s
And, circum-radius=abc/4Δ
Ratio=4Δ²/abcs=8Δ²/2ycs
Now, Δ=1/2ab.sinC
cosC=(a²+b²-c²)/2ab=(x²-c²)/2y=y-c²/2y=1/2
=>C=60°
So,sinC=√3/2
Δ=1/2y.√3/2=√3/4y
Substituting this in the ratio expression gives
Ratio=8(√3/4y)²/2yc(x+c) = 8(3/16y²)/2yc(x+c) = 3y²/2yc(x+c) = 3y/2c(x+c)
Let's use the formula for the area of a triangle:
Δ = √(s(s-a)(s-b)(s-c))
where s is the semi-perimeter, s = (a+b+c)/2
In this case, a+b = x, and ab = y
Also, x² - c² = y
Let's use Heron's formula for the area of a triangle:
Δ = √(s(s-a)(s-b)(s-c))
where s = (a+b+c)/2 = (x+c)/2
Δ = √(((x+c)/2)((x+c)/2 - a)((x+c)/2 - b)((x+c)/2 - c))
Δ = √(((x+c)/2)((x+c-2a)/2)((x+c-2b)/2)((x-c)/2))
We have a+b = x and ab = y
(a+b)² = a² + b² + 2ab = x²
a² + b² = x² - 2y
From the cosine rule:
cosC = (a²+b²-c²)/(2ab) = (x²-2y-c²)/(2y) = (y-c²)/(2y) = 1/2
Thus, C = 60°
Δ = (1/2)ab sinC = (1/2)y sin60° = (√3/4)y
Inradius r = Δ/s = (√3/4)y/((x+c)/2) = (√3/2)y/(x+c)
Circumradius R = abc/(4Δ) = (yc)/(4((√3/4)y)) = c/(√3)
Ratio r/R = ((√3/2)y/(x+c))/(c/√3) = (3y)/(2c(x+c))
Hence, Option A