[JEE Advanced 2004] Find the centre and radius of the circle $\left|\dfrac{z-\alpha}{z-\beta}\right|=k$ ($k\ne1$), where $z=x+iy$, $\alpha,\beta$ complex.
Find the centre and radius of the circle $\left|\dfrac{z-\alpha}{z-\beta}\right|=k$ ($k\ne1$), where $z=x+iy$, $\alpha,\beta$ complex.
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 1mo ago
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Answer: Centre $\dfrac{\alpha-k^2\beta}{1-k^2}$, radius $\dfrac{k\,|\alpha-\beta|}{|1-k^2|}$.
$|z-\alpha|^2=k^2|z-\beta|^2$ expands to an Apollonius circle with the stated centre and radius.
JEE Advanced 2004 · Complex Numbers — verified solution by the Vidaara Team.
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