Question

If $4l^2-5m^2+6l+1=0$ then the centre and radius of the circle which have $lx+my+1=0$ is a tangent at

• A. $(0,4);\sqrt{5}$
• B. $(4,0);\sqrt{5}$
• C. $(0,3);\sqrt{5}$
• D. $(3,0);\sqrt{5}$

Answer 1

Answer: (D)

$(3,0);\sqrt{5}$

$4l^2-5m^2+6l+1=0$
$\Rightarrow 4l^2+6l+1=5m^2$
Add $5l^2$ both sides, we get
$\Rightarrow 4l^2+5l^2+6l+1=5m^2+5l^2$
$\Rightarrow 9l^2+6l+1=5(m^2+l^2)$
$\Rightarrow {(3l+1)}^2=5(m^2+l^2)$
$\Rightarrow \frac{|3l+1|}{\sqrt{(l^2+m^2)}}=\sqrt{5}$
$\therefore$ Centre$\equiv (3,0)$ and radius$\equiv \sqrt{5}$