# 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?

## 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?

Jun 27, 2018

$a : b = 3 \pm 2 \sqrt{2}$

#### 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 $y = x$, as seen below: The common chord is going to be $P Q$, where $Q$ is the symmetric point (a.k.a. the reflection) of $P$ with respect to the line $y = x$, hence $Q$ has coordinates $\left(b , a\right)$.

In order for the common chord to be the maximised, it has to be the diameter of the circle where $P$ and $Q$ are on its "high ends": In our particular example, it almost overlaid one of the original circles.

Anyway, the center of this new circle is the midpoint of $P Q$, which we will denote $L$.

$L \left(\frac{a + b}{2} , \frac{a + b}{2}\right)$

The equation of the circle is:

${\left(x - \frac{a + b}{2}\right)}^{2} + {\left(y - \frac{a + b}{2}\right)}^{2} = {\left(\frac{P Q}{2}\right)}^{2}$

The lenght of $P Q$ is, as calculated by you, $\sqrt{2} | a - b |$.

After some arithmetic, we reach the equation

(x−a)(x−b)+(y−a)(y−b)=0

The circle with diameter $P Q$ must touch the axes as well, hence the points $\left(\left(a + b\right) \text{/} 2 , 0\right)$ and $\left(0 , \left(a + b\right) \text{/} 2\right)$ must be on it as well.

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

${\left(a + b\right)}^{2} = 8 a b$

${a}^{2} + 2 a b + {b}^{2} = 8 a b$

${a}^{2} - 6 a b + {b}^{2} = 0$

Divide both sides by ${b}^{2}$:

${a}^{2} / {b}^{2} - 6 \frac{a}{b} + 1 = 0$

${\left(\frac{a}{b}\right)}^{2} - 6 \left(\frac{a}{b}\right) + 1 = 0$

This is a quadratic equation with the indeterminate the ratio of $a$ and $b$, exactly what we want!

The solutions of this equation are

$a : b = 3 \pm 2 \sqrt{2}$

These are the ratios we desire.