Question

The equation of the circle whose diameter is common chord to the circles
$x^2+y^2+2ax+c=0$ and $x^2+y^2+2by+c=0$ is

• A. $x^2+y^2-\frac{2a{b}^2}{a^2+b^2}x+\frac{2a^{2}b}{a^2+b^2}y+c=0$
• B. $x^2+y^2-\frac{2a{b}^2}{a^2+b^2}x-\frac{2a^{2}b}{a^2+b^2}y+c=0$
• C. $x^2+y^2+\frac{2a{b}^2}{a^2+b^2}x+\frac{2a^{2}b}{a^2+b^2}y+c=0$
• D. $x^2+y^2+\frac{2a{b}^2}{a^2+b^2}x-\frac{2a^{2}b}{a^2+b^2}y+c=0$

Answer 1

Answer: (C)

$x^2+y^2+\frac{2a{b}^2}{a^2+b^2}x+\frac{2a^{2}b}{a^2+b^2}y+c=0$

Let $S_1\equiv x^2+y^2=2ax+c=0$
and $S_2\equiv x^2+y^2+2by+c=0$

Equation of common chord as a diameter of the third circle
$S_1-S_2=0$
$ax-by=0$
$y=\frac{ax}{b}$ ...... (i)

On putting the value of y in Eq. (i), we get
$x^2+\left(\frac{a^2{x}^2}{b^2}\right)+2ax+c=0$
$(a^2+b^2)x^2+2a{b}^{2}x+c{b}^2=0$

Let $x_1$ and $x_2$ be the roots of the equation.
$\therefore x_1+x_2=\frac{-2a{b}^2}{a^2+b^2}$ and $x_1{x}_2=\frac{c{b}^2}{a^2+b^2}$

Similarly,

$(a^2+b^2)y^2+2b{a}^{2}y+c{a}^2=0$

Let $y_1$ and $y_2$ be the roots of the equation
$\therefore y_1+y_2=\frac{-2a^{2}b}{a^2+b^2}$
and $y_1{y}_2=\frac{c{a}^2}{a^2+b^2}$

Now, the required equation of third circle
$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$
$x^2-(x_1+x_2)x+x_1{x}_2+y^2-(y_1+y_2)y+y_1{y}_2=0$
$x^2+y^2+\frac{2a{b}^2}{a^2+b^2}\cdot x+\frac{2b{a}^2}{a^2+b^2}\cdot y+\frac{c{b}^2}{a^2+b^2}+\frac{c{a}^2}{a^2+b^2}=0$

$\therefore x^2+y^2+\frac{2a{b}^2}{a^2+b^2}\cdot x+\frac{2a^{2}b}{a^2+b^2}\cdot y+\frac{c(a^2+b^2)}{(a^2+b^2)}=0$
$\Rightarrow x^2+y^2+\frac{2a{b}^2}{(a^2+b^2)}\cdot x+\frac{2a^{2}b}{(a^2+b^2)}\cdot x+\frac{2a^{2}b}{(a^2+b^2)}\cdot y+c=0$