Finding the range of product of two distances.

Question

Two circles with the same center their radii are $7$ and $12$ cm the cord $AD$ is drawn in the greater circle to intersect the smallest circle in $B$ and $C$ , show that the value of the product $$AB.AD \in [120 ,190]$$

Answer 1

Let us denote the common center of the circles by O. From the common center O draw the segment AH perpendicular to AD where H is on AD. Note that H is the midpoint of AD. Let r be the length of AH. From the right triangle AOH we find $$AH=\sqrt {144-r^2}$$ Therefore $$ AD = 2 \sqrt {144-r^2}$$ Note that $AB =AH-BH$ and $BH= \sqrt {49-r^2}$.

Thus $$AB = \sqrt {144-r^2}-\sqrt {49-r^2}$$ We get $$AB\times AD =(\sqrt {144-r^2}-\sqrt {49-r^2})(2 \sqrt {144-r^2})$$ The range of $r$ is $[0,7]$ which results in $$120\le AB\times AD \le 190$$

Answer 2

AB attain minima and AD attain maxima when AD passes through centre while AB attain maximum and AD attain minima when AD is tangent to smaller circle. So in first case $AB=5$ and $AD=24$ while in the second case $AB=\sqrt {95}$ and $AD=2\sqrt {95}$

$$\Rightarrow AB\in[5,\sqrt{95}]$$

And $$AD\in[2\sqrt{95},24]$$

So by basic intuition we get the range of product as $$AD.AB\in[120, 190]$$