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If the inverse point of the point $(3,2)$ with respect to the circle $\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}+4 \mathrm{y}-4=0$ is $(l, \mathrm{~m})$ then $(2 l+19 \mathrm{~m})=$
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$0$
Given circle is $x^2+y^2-2 x+4 y-4=0$
$\therefore$ Centre $=(1,-2)$ and radius $=\sqrt{1+4+4}=3$
We have inverse of point $P(\alpha, \beta)$ w.r. to circle whose centre $(h, k)$ and radius $\mathrm{r}$ is
$$
x=\lambda(\alpha-h)+h, y=\lambda(\beta-k)+k,
$$
$$
\text { where } \lambda=\frac{r^2}{(\alpha-h)^2+(\beta-k)^2}
$$
$$
\therefore \lambda=\frac{9}{(3-1)^2+(2+2)^2}=\frac{9}{20}
$$
$\therefore$ Inverse point $(l, m)$
$$
=\left[\frac{9}{20}(2)+1, \frac{9}{20}(4)-2\right]=\left[\frac{38}{20},-\frac{4}{20}\right]
$$
Now, $2 l+19 m=2\left(\frac{38}{20}\right)+19\left(\frac{-4}{20}\right)=0$
$\therefore$ Centre $=(1,-2)$ and radius $=\sqrt{1+4+4}=3$
We have inverse of point $P(\alpha, \beta)$ w.r. to circle whose centre $(h, k)$ and radius $\mathrm{r}$ is
$$
x=\lambda(\alpha-h)+h, y=\lambda(\beta-k)+k,
$$
$$
\text { where } \lambda=\frac{r^2}{(\alpha-h)^2+(\beta-k)^2}
$$
$$
\therefore \lambda=\frac{9}{(3-1)^2+(2+2)^2}=\frac{9}{20}
$$
$\therefore$ Inverse point $(l, m)$
$$
=\left[\frac{9}{20}(2)+1, \frac{9}{20}(4)-2\right]=\left[\frac{38}{20},-\frac{4}{20}\right]
$$
Now, $2 l+19 m=2\left(\frac{38}{20}\right)+19\left(\frac{-4}{20}\right)=0$
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