The line 3x - 2y = k meets the circle {x^2} + | vedclass

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Question

(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 = \fr

Vedclass · Accepted Answer

$52{r^2}$

Vedclass · Answer

(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}} ight)}^2}} $

{From $(ii)$, $a = 2r$}

$ = \frac{{{k^2}}}{4} = 4{r^2} 	imes \frac{{13}}{4}$,

$	herefore $ ${k^2} = 52{r^2}.$

The line $3x - 2y = k$ meets the circle ${x^2} + {y^2} = 4{r^2}$ at only one point, if ${k^2}$=

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