Question

The line $y=mx$ intersects the circle $x^2+y^2-4x-4y=0$ and $x^2+y^2+6x-8y=0$ at points $A$ & $B$ (points being other than the origin). The range of $m$ such that the origin divides $AB$ internally is

• A. $m>\frac{4}{3}$ or $m<-2$
• B. $m>-1$
• C. $-2<m<\frac{4}{3}$
• D. $-1<m<\frac{3}{4}$

Answer 1

The correct option is D $-1<m<\frac{3}{4}$


$O$ divides $AB$ internally in the ratio $\lambda :1$
$\lambda >0$
$A(x_1,y_1),B(x_2,y_2),y=mx$
$\Rightarrow A(x_1,m{x}_1),B(x_2,m{x}_2)$
Section formula : $(0,0)=(\frac{\lambda x_2+x_1}{\lambda +1},\frac{\lambda m{x}_2+m{x}_1}{\lambda +1})$
$\Rightarrow x_1=-\lambda x_2$
Or $\lambda =\frac{-x_1}{x_2}>0\Rightarrow \frac{x_1}{x_2}<0$

$x_1^2+m^2{x}_1^2-4x_1-4m{x}_1=0;x_2^2+m^2{x}_2^2+6x_2-8m{x}_2=0$

$\Rightarrow x_1=\frac{4(1+m)}{1+m^2};\Rightarrow x_2=\frac{2(4m-3)}{1+m^2}$
$\because x_1.x_2<0$
$\Rightarrow (1+m)(4m-3)<0$
Critical points $m=-1,\frac{3}{4}$
$\Rightarrow -1<m<\frac{3}{4}$