# P(a, b) is a point in the first quadrant. Circles are drawn through P touching the coordinate axes, such that the length of common chord of these circles is maximum, find the ratio a : b?

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while solving the question, i realized that there will always be two such circles that can be drawn through P and satisfying the above criterion, with radius:

r1 = a+b+(2ab)^0.5 and r2 = a+b-(2ab)^0.5

the length of the common chord then turned out to be something like

L = root2*modulus(a-b)

if the question were to minimize this length, a : b would have easily been 1. but for maximising, woudn't the answer be infinity?

while solving the question, i realized that there will always be two such circles that can be drawn through P and satisfying the above criterion, with radius:

r1 = a+b+(2ab)^0.5 and r2 = a+b-(2ab)^0.5

the length of the common chord then turned out to be something like

L = root2*modulus(a-b)

if the question were to minimize this length, a : b would have easily been 1. but for maximising, woudn't the answer be infinity?

##### 1 Answer

#### Explanation:

In order for a circle to be touching both coordinate axes in the first quadrant, it must have its center located on the line

The common chord is going to be

In order for the common chord to be the maximised, it has to be the diameter of the circle where

In our particular example, it almost overlaid one of the original circles.

Anyway, the center of this new circle is the midpoint of

The equation of the circle is:

The lenght of

After some arithmetic, we reach the equation

The circle with diameter **must** touch the axes as well, hence the points

If we substitute the coordinates of one of these points (doesn't matter which; it gives us the same answer) into the circle equation we get the condition

Divide both sides by

This is a quadratic equation with the indeterminate the ratio of

The solutions of this equation are

These are the ratios we desire.