Question

$A$and $B$ are two points on the x-axis and y-axis respectively. Two circles are drawn passing through the origin and having centre at $A$ and $B$

• A. Equation of the common chord is $ax-by=0$
• B. mid-point of the common chord is $(\frac{a{b}^2}{a^2+b^2},\frac{a^{2}b}{a^2+b^2})$
• C. $AB$ bisects the common chord.
• D. $AB$ is perpendicnlar to the common chord.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Equation of the common chord is $ax-by=0$
B mid-point of the common chord is $(\frac{a{b}^2}{a^2+b^2},\frac{a^{2}b}{a^2+b^2})$
C $AB$ bisects the common chord.
D $AB$ is perpendicnlar to the common chord.

Let $OA=a$ and $OB=b\left(Fig\right),$ so that the equation of the line $AB$ is $\frac{x}{a}+\frac{y}{b}=1$.

Equations of the circles passing through the origin and having centres at $A(a,0)$ and $B(0,b)$ are, respectively,
$x^2-2ax+y^2=0$ ....(1)

and $x^2-2by+y^2=0$ ...(2)
so that the equation of /he common choro $OP$ is

$ax-by=0$ ...(3)

which is perpendicular to $AB$
This has one end at the origin $O$ and the other end $P$ is given by solving $\left(2\right)$ and $\left(3\right).$
${\left(\frac{b}{a}y\right)}^2-2by+y^2=0$
$\Rightarrow y=\frac{2a^{2}b}{a^2+b^2}\Rightarrow x=\frac{2a{b}^2}{a^2+b^2}$

Therefore, the mid-point of the common chord $OP$ is
$(\frac{a{b}^2}{a^2+b^2},\frac{a^{2}b}{a^2+b^2})$
which lies on $AB$