Question

If $4l^2-5m^2+6l+1=0$, then the line $lx+my+1=0$ touches the circle $x^2+y^2-6x+k=0$. The value of $k$ is:

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

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

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

If possible let line (2) touches the circle whose center is $(\alpha ,\beta )$ and radius is $a$, then

$\frac{|l\alpha +m\beta +1|}{\sqrt{l^2+m^2}}=a\Rightarrow {(l\alpha +m\beta +1)}^2=a^2(l^2+m^2)\Rightarrow l^2{\alpha }^2+m^2{\beta }^2+1+2lm\alpha \beta +2l\alpha +2m\beta =a^2{l}^2+a^2{m}^2$

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

Comparing (1) and (3), we get

${\alpha }^2-a^2=4$ ...(4)

${\beta }^2-a^2=4$ ...(5)

$2\alpha =6$ ...(6)

$2\beta =0$ ...(7)

and $2\alpha \beta =0$ ...(8)

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

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

$a^2=32-5=5\Rightarrow a=\sqrt{5}$

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

$\therefore$ Equation of the required circle is

${(x-3)}^2+{(y-0)}^2={\left(\sqrt{5}\right)}^2\Rightarrow x^2+y^2-6x+4=0\Rightarrow k=4$