Two circles having radii $7$cm & $19$ cm are separated by a distance of $22$ cm between their centers. If they are intersecting each other at two points $P$ & $Q$ then what will be the length of the common chord PQ?
a.) $\frac{9\sqrt{205}}{13}$
b.) $\frac{8\sqrt{255}}{11}$
c.) $\frac{9\sqrt{233}}{13}$
d.) $\frac{8\sqrt{155}}{11}$
I tried this by assuming one circle to be centered at the origin but it's creating utter mess. I am 12th grade. Thanks for your help!
On
Connect the centers and the intersection points.
Using this enter link description here you find the area to be $4 \sqrt{255}$. $22h/2$ will also be the area. So you find $h$ to be $4 \sqrt{255}/11$ But the chord is $2h$. So final answer
$$\frac{8 \sqrt{255}}{11}$$
$$7^2-x^2=19^2-(22-x)^2$$ $$x=\frac{43}{11}$$ so $$h^2=7^2-\frac{43^2}{11^2}=\frac{4080}{121}$$ $$h=\frac{4\sqrt{255}}{11}$$ hence the Chord $PQ$ = $2h$
$$PQ=\frac{8\sqrt{255}}{11}$$