Question

$L_1$ is a line intersecting $x$ and $y$ axes at $A(a,0)$ and $B(0,b).$ $L_2$ is a line perpendicular to $L_1$ intersecting $x$ and $y$ axes at $C$ and $D$ respectiveley. What is the condition for the common chord of the circles with $BD$ and $AC$ as diameters to pass through the point $(a,b)?$

• A. $a=2b$
• B. $2a=b$
• C. $a=-b$
• D. $a=-2b$

Answer 1

The correct option is C $a=-b$

$B(0,b),D(0,-a\alpha )$
$x^2+(y-b)(y+a\alpha )=0$(Circle with $BD$ as diameter)
$x^2+y^2+(a\alpha -b)y-ab\alpha =0$
$A(a,0),C(\alpha b,0)$
($AC$ as diameter)
$(x-a)(x-\alpha b)+y^2=0\Rightarrow x^2+y^2-(a+\alpha b)x+ab\alpha =0$
Two equations are
$x^2+y^2+(a\alpha -b)y-ab\alpha =0$
$x^2+y^2-(a+\alpha b)x+ab\alpha =0$
On subtracting the above equations , we get
$(a+\alpha b)x+(a\alpha -b)y-2ab\alpha =0$ (equation of chord)
Since, the chord passes through $(a,b)$ therefore, put $x=a,y=b$
$\Rightarrow a^2=b^2$
$\Rightarrow a=\pm b$