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Two Circles passing through center of other circle. Find length of common chord when length of chords for each circle which are formed by straight line passing to one intersection point are 3 and 7 respectively.
Step 1:
First you have to observe, both circles are passing through center of other circle.
⇒ Radius of both circles are same.
Let radius = r
Step 2:
Draw all possible lines with length r from center of each circle to given external point in previous diagram.
For each circle, 4 lines can be drawn as shown in figure.
in the above figure, all black dashed lines are of length r.
Now observe Triangle PQD
All lines are of length r.
⇒ Triangle PQD is equilateral triangle
⇒ ∠PDC = 60
Step 3:
Let ∠PDC = θ
⇒ ∠QDB = 120-θ
Now apply cosine rule on ∠PDC and ∠QDB.
\[cos\theta=\frac{9+r^2-r^2}{2\times3\times r}=\frac{3}{2r}\]
\[cos(120-\theta)=\frac{49+r^2-r^2}{2\times7\times r}=\frac{7}{2r}\]
Now expand cos(120-θ)
\[cos(120-\theta)=\frac{7}{2r}\]
\[\Rightarrow cos120.cos\theta+sin120.sin\theta=\frac{7}{2r}\]
\[\Rightarrow \frac{-1}{2}.\frac{3}{2r}+\frac{\sqrt{3}}{2}.\frac{\sqrt{4r^2-9}}{2r}=\frac{7}{2r}\]
\[\Rightarrow -3+\sqrt{3}\sqrt{4r^2-9}=14\]
\[\Rightarrow \sqrt{3}\sqrt{4r^2-9}=17\]
\[\Rightarrow 3(4r^2-9)=289\]
\[\Rightarrow 12r^2-27=289\]
\[\Rightarrow 12r^2=316\]
\[\Rightarrow r^2=\frac{316}{12}\]
\[\Rightarrow r^2=\frac{79}{3}\]
Step4:
Now apply cosine rule on ∠APD. Where ∠APD = 120.
\[cos120 = \frac{r^2+r^2-AD^2}{2\times r\times r}\]
\[\Rightarrow \frac{-1}{2} = \frac{2r^2-AD^2}{2r^2}\]
\[\Rightarrow AD^2=3r^2\]
\[\Rightarrow AD^2=3.\frac{79}{3}\]
\[\Rightarrow AD^2=79\]
\[\Rightarrow AD=\sqrt{79}\]
So length of AD=√79
