Question

Consider the relation $4l^2-5m^2+6l+1=0$, where $l,m\in R$, then the line $lx+my+1=0$ touches a fixed circle whose tangents $PA$ and $PB$ are drawn to the above fixed circle from the points $P$ on the line $x+y-1=0$. Then chord of contact $AB$ passes through the fixed point

• A. $(\frac{1}{2},\frac{-5}{2})$
• B. $(\frac{1}{3},\frac{4}{3})$
• C. $(\frac{-1}{2},\frac{3}{2})$
• D. $(\frac{-1}{2},\frac{4}{3})$

Answer 1

The correct option is A $(\frac{1}{2},\frac{-5}{2})$

Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0\left(i\right)$
The line $lx+my+1=0$ will touch circle $\left(i\right)$ if the length of perpendicular from the centre $(-g,-f)$ of the circle on the line is equal to its radius, i.e.
$\frac{|-gl-mf+1|}{\sqrt{l^2+m^2}}=\sqrt{g^2+f^2-c}$
$(gl+mf-1{)}^2=(l^2+m^2\left)\right(g^2+f^2-c)$
or $(c-f^2)l^2+(c-g^2)m^2-2gl-2fm+2gflm+1=0\left(ii\right)$
But the given condition of tangency is $4l^2-5m^2+6l+1=0\left(iii\right)$
Comparing $\left(ii\right)$ and $\left(iii\right)$, we get
$c-f^2=4,c-g^2=-5,-2g=6,-2f=0,2gf=0$
Solving, we get
$f=0,g=-3,c=4$
Substituting these values in $\left(i\right)$, the equation of the circle is $x^2+y^2-6x+4=0$. Any point on the line $x+y-1=0$ is $(t,1-t),t\in R$

The chord of contact generated by this point for the circle is $tx+y(1-t)-3(t+x)+4=0$ or $t(x-y-3)+(-3x+y+4)=0$, which is concurrent at the point of intersection of the lines $x-y-3=0$ and $-3x+y+4=0$ for all values of $t$. Hence, the lines are concurrent at $(\frac{1}{2},-\frac{5}{2})$