Q19. - Geometry

Question

$\odot P$ and $\odot Q$ have radii $5$ and $7$ and $PQ=6$. Find the length of the common chord $\overline{AB}$. $APBQ$ is a kite and $\overline{PQ}$ is he perpendicular bisector of $\overline{AB}$. Let be the intersection of $\overline{PQ}$ and $\overline{AB}$. Let $PN=x$ and $AN=y$. Write two equations in term of x and y.

Step-by-Step Solution

Answer

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

Step 1. Given information.

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

Step 2. 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 $\Delta ANQ$.

Step 3. Solution.

 $\begin{array}{c}LetPN=x,\\ AN=y,\\ NQ=6-x\end{array}$

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

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

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

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

 By (i) and (ii), 

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

So, by equation (i),

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

$PQ$ bisects chord $AB$.

Therefore, 

 $\begin{array}{l}AB=2AN\\ AB=4\sqrt{6.}\end{array}$