Chapter 10.1, Problem 25PSC

Solution

To Prove : The length of common chord.

  $24$

Given information : $\begin{array}{l}\text{Two circles intersect and have a common chord}\text{. Radii of circles are}\\ \text{13}\text{cm and 15 cm distance between there centres is 14 cm}\text{.}\end{array}$

Explanation : Let the circle centres are A and B common chord is CO. Draw radii AC and BC.

  $\begin{array}{l}\text{And, }AC=13,BC=15,AB=14\\ \text{Now, In }\Delta ABC,\text{ using cosine rule}\text{.}\\ \cos A=\frac{A{B}^2+A{C}^2-B{C}^2}{2AB\cdot AC}\\ \cos A=\frac{{14}^2+{13}^2-{15}^2}{2\cdot 14\cdot 13}\\ \cos A=\frac{146+169-225}{2\cdot 14\cdot 13}=\frac{140}{364}\\ \cos A=\frac{5}{13}\end{array}$   

We know that line through their centres is perpendicular bisector of common chord CD. It means,

  $\begin{array}{l}\text{(i)}\text{If we find CE, double of }CE\text{ is }CD\text{.}\\ \text{(ii)}ABC\text{ is a right triangle}\text{.}\\ \text{In triangle }ABC,\sin A=\frac{CE}{AC}\\ CE=AC\cdot \sin A\\ CE=13\sin A\\ \text{Now, we use the identity}\\ {\sin }^{2}A+{\cos }^{2}A=1\\ {\sin }^{2}A=1-{\cos }^{2}A\\ \text{Substitute }\frac{\text{5}}{\text{13}}\text{ for}\cos A\\ {\sin }^{2}A=1-{\left(\frac{5}{13}\right)}^2\\ {\sin }^{2}A=1-\frac{25}{169}\\ {\sin }^{2}A=\frac{169-25}{169}\\ {\sin }^{2}A=\frac{144}{169}\\ \sin A=\sqrt{\frac{144}{169}}\\ \sin A=\frac{12}{13}\\ \text{Now, substitute that in}\\ CE=13\cdot \sin A\\ CE=13\cdot \frac{12}{13}=12\\ \text{Therefore, }CD=2CE\\ CD=2\cdot 12\\ CD=24\end{array}$