Question

If $4l^2-5m^2+61+1=0$ and the line $1x+my+1=0$ touches a fixed circle then that circle

• A. has centre $(4,0)$
• B. has radius $2$
• C. has radius $5$
• D. pass

Answer 1

Answer: (C)

has radius $5$

Given: $4l^2-5m^2+6l+1=0$ .......(1)

and line is $lx+my+1=0$ .......(2)

let line $2$ touch the circle whose center is $(\alpha ,\beta )$ and radius is a than

$\frac{|la+m\beta +1|}{\sqrt{l^2+m^2}}=a$

$=(la+m\beta +1{)}^2=a^2(l^2+m^2)$

$\Rightarrow l{\alpha }^2+m^2{\beta }^2+1+2lm\alpha \beta +2l\alpha +2m\beta =a^2{l}^2+{\alpha }^2{m}^2$

$\Rightarrow (l^2-{\alpha }^2)l^2+({\beta }^2-{\alpha }^2)m^2+2lm\alpha \beta +2\alpha l+2\beta m-1)=0$....(3)

comparing (1) and (3) we get

${\alpha }^2-a^2=\mathrm{4....}\left(4\right)$ and $\beta -a^2=-5....\left(5\right)$

$2a=6.....\left(6\right),2\beta =\mathrm{0.....}\left(7\right)$ and $2\alpha \beta =0.....\left(8\right)$

from (6) we get $\alpha =3$ and from (7) we get $\beta =0$

putting the value of $\alpha$ in (4) we get

$a^2=3^2-4=5$

$\Rightarrow a=\sqrt{5}$

Thus, the center of the required circle is $(3,0)$ and radius is $\sqrt{5}$

Hence the equation of the required circle is

$(x-\alpha {)}^2+(y-\beta {)}^2=a^2$

$\Rightarrow (x-3{)}^2+(y-0{)}^2=5$

$\Rightarrow (x-3{)}^2+y^2=5$