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Consider a circle passing through the origin and the points $(\mathrm{a}, \mathrm{b})$ and $(-b,-a)$.
On which line does the centre of the circle lie?
Options:
On which line does the centre of the circle lie?
Solution:
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Verified Answer
The correct answer is:
$x+y=0$
Suppose; $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$ is the eq. of the
circle. Since; it passes through $(0,0) ;(a, b) \&(-b,-a)$
$\therefore \mathrm{C}=0$
$a^{2}+b^{2}+2 g a+2 f b=0$
$\therefore \mathrm{a}^{2}+\mathrm{b}^{2}-2 \mathrm{gb}-2 \mathrm{fb}=0$
on solving:
$\mathrm{g}=-\mathrm{f}$
$\therefore$ centre $=(-\mathrm{g},-\mathrm{f})$ or $(+\mathrm{f},-\mathrm{f})$
$\therefore$ from options:
$\mathrm{x}+\mathrm{y}=0$ is the line which passes through $(f,-f)$
circle. Since; it passes through $(0,0) ;(a, b) \&(-b,-a)$
$\therefore \mathrm{C}=0$
$a^{2}+b^{2}+2 g a+2 f b=0$
$\therefore \mathrm{a}^{2}+\mathrm{b}^{2}-2 \mathrm{gb}-2 \mathrm{fb}=0$
on solving:
$\mathrm{g}=-\mathrm{f}$
$\therefore$ centre $=(-\mathrm{g},-\mathrm{f})$ or $(+\mathrm{f},-\mathrm{f})$
$\therefore$ from options:
$\mathrm{x}+\mathrm{y}=0$ is the line which passes through $(f,-f)$
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