Radius Ratio #1

Imgur

Since the circles of radii a and b lie in the first quadrant and are touching both axes, the two centers would be (a,a) and (b,b) Their centers C1, C2 along with the point of orthogonal intersection T would form a right angle triangle, because they intersect orthogonally. $a^2 + b^2 = D^2$ Where D = distance between C1 and C2, which can also be written as $D = \sqrt{2}b - \sqrt{2}a$ Equating the two expressions for D we get $\frac{b}{a} = 2 + \sqrt{3}$

Alright this problem is not that messy

As the circles touch both the coordinate axes their centers will be same the radius i.e.

(a , a) And (b, b)

Equation of circle can be written as

$x^{2} + y^{2} + 2gx + 2fy + c = 0$

$radius = \sqrt{ g^{2} + f^{2} - c}$

Centre is (-g, -f)

here we have equation as

$x^{2} + y^{2} - 2ax - 2ay + a^{2} = 0$

And

$x^{2} + y^{2} - 2bx - 2by + b^{2} = 0$

For circle which are orthogonal we have condition

$2g_1g_2 + 2f_1f_2 = c_1 + c_2$

Substituting values we will have

2ab + 2ab = $a^{2} + b^{2}$

Dividing by 'ab' we have

$\frac{b}{a} + \frac{a}{b} = 4$

Let $\frac{b}{a} = t$

We will have

$\displaystyle t^{2} - 4t + 1 = 0$

Solving for 't' we will have

$\frac{b}{a} = \boxed{2 + \sqrt{3}}$