Chapter 9.1, Problem 19WE

To determine

To find:The length of the common chord $AB$ .

Answer to Problem 19WE

The length of the common chord $AB$ is $4\sqrt{6}$ units.

Explanation of Solution

Giveninformation: Two circles P and Q with radii 5 and 7 and $PQ=6$ .

Concept used:Join $AP$ and $AQ$ , so that these are the radii of two circles. Further by property, any perpendicular from center to a chord of a circle is always perpendicular on it. So, apply Pythagorean theorem and then we get two right triangles $\Delta ANP$ and $\Delta ANQ$ .

  Calculation:Let $PN=x,AN=y\Rightarrow NQ=6-x$

So, in right triangle $\Delta ANP$ , using Pythagorean theorem,

  $\begin{array}{l}A{P}^2=A{N}^2+P{N}^2\\ \Rightarrow 5^2=y^2+x^2\\ \Rightarrow y^2+x^2=25\mathrm{......}(i)\end{array}$

Again in right triangle $\Delta ANQ$ , using Pythagorean theorem,

  $\begin{array}{l}A{Q}^2=A{N}^2+Q{N}^2\\ \Rightarrow 7^2=y^2+{(}^6\\ \Rightarrow y^2+x^2-12x+36=47\mathrm{......}(ii)\end{array}$

(It is because ${(a-b)}^2=a^2+b^2-2ab$ )

By (i) and (ii),

  $\begin{array}{l}25-12x+36=49\\ \Rightarrow 61-12x=49\\ \Rightarrow -12x=49-61=-12\\ \Rightarrow x=1\end{array}$

So, by equation (i),

  $\begin{array}{l}{1}^2+y^2=25\\ y^2=24\\ \Rightarrow y=AN=\sqrt{24}=2\sqrt{6}\end{array}$

Now as PQ bisects chord $AB$ also, so, $AB=2AN=4\sqrt{6}$ .

Conclusion:So, the length of the chord $AB$ = $4\sqrt{6}$ units.