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If the centre of the circle passing through the origin is $(3,4)$, then the intercepts cut off by the circle on $\mathrm{x}$ -axis and $\mathrm{y}$ -axis respectively are $\quad[2015-I I]$
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The correct answer is:
6 unit and 8 unit
Equation of circle having radius $\mathrm{r}$ and centre $(3,4)$ is $=(x-3)^{2}+(y-4)^{2}=r^{2}$
if it is passing through $(0,0)$ $\therefore(0-3)^{2}+(0-4)^{2}=\mathrm{r}^{2}$
$\Rightarrow \mathrm{r}^{2}=25$
equation of circle is
$$
(x-3)^{2}+(y-4)^{2}=25
$$
putting $\mathrm{y}=0$ $\therefore \mathrm{x}=6$ unit $=$ interception $\mathrm{x}$ -axis
intercept on y axis (putting $\mathrm{x}=0$ ) is $y-8$ unil
if it is passing through $(0,0)$ $\therefore(0-3)^{2}+(0-4)^{2}=\mathrm{r}^{2}$
$\Rightarrow \mathrm{r}^{2}=25$
equation of circle is
$$
(x-3)^{2}+(y-4)^{2}=25
$$
putting $\mathrm{y}=0$ $\therefore \mathrm{x}=6$ unit $=$ interception $\mathrm{x}$ -axis
intercept on y axis (putting $\mathrm{x}=0$ ) is $y-8$ unil
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