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10-1.Circle and System of Circles
medium
The line $3x - 2y = k$ meets the circle ${x^2} + {y^2} = 4{r^2}$ at only one point, if ${k^2}$=
A
$20{r^2}$
B
$52{r^2}$
C
$\frac{{52}}{9}{r^2}$
D
$\frac{{20}}{9}{r^2}$
Solution
(b) Equation of line is
$3x – 2y = k$……$(i)$
Circle is ${x^2} + {y^2} = 4{r^2}$…..$(ii)$
Equation of line can be written as
$y = \frac{3}{2}x – \frac{k}{2}$
Here, $c = – \frac{k}{2},\,m = \frac{3}{2}$
Now the line will meet the circle at one point, if $c = \pm a\sqrt {1 + {m^2}} $
$ = \frac{{ – k}}{2} = \pm (2r)\,\sqrt {1 + {{\left( {\frac{3}{2}} \right)}^2}} $
{From $(ii)$, $a = 2r$}
$ = \frac{{{k^2}}}{4} = 4{r^2} \times \frac{{13}}{4}$,
$\therefore $ ${k^2} = 52{r^2}.$
Standard 11
Mathematics
