Equation to the circle whose one of the diameters is the common chord of
$(x - a)^{2} + y^{2} = a^{2}, x^{2} + (y - b)^{2} = b^{2}$ is

(a^{2} + b^{2}) (x^{2} + y^{2}) = 2 a b (b x + a y)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(a^{2} + b^{2}) (x^{2} + y^{2}) = 2 a b (a x + b y)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} = \frac{2 a b}{(a^{2} + b^{2}) (a x - b y)}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} = \frac{a b}{(a^{2} + b^{2}) (a x + b y)}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $(a^{2} + b^{2}) (x^{2} + y^{2}) = 2 a b (b x + a y)$
Equation of common chord,
$s_{1} - s_{2} = 0$
$2 a x - 2 b y = 0$
$a x = b y . . . . . (1)$
$x = \frac{b y}{a}$
Put $x = \frac{b y}{a}$ in $x^{2} - (y - b)^{2} = b^{2}$
$\frac{b^{2} y^{2}}{a^{2}} + y^{2} - 2 b y = 0$
$y (a^{2} + b^{2}) - 2 b^{2} = 0$
$y = \frac{2 b a^{2}}{(a^{2} + b^{2})}$ and $x = \frac{2 a b^{2}}{(a^{2} + b^{2})}$
Equation of circle whose diametric ends are $(0, 0)$ and $(\frac{2 a b^{2}}{a^{2} + b^{2}}, \frac{2 a^{2} b}{a^{2} + b^{2}})$ is
$x (x - \frac{2 a b^{2}}{a^{2} + b^{2}}) + y (y - \frac{2 a^{2} b}{a^{2} + b^{2}}) = 0$
$\Rightarrow (a^{2} + b^{2}) (x^{2} + y^{2}) = 2 a b (b x + a y)$

Suggest Corrections

Similar questions

The length of common chord of the circles $(x - a)^{2} + y^{2}$ = $a^{2}$ and $x^{2} + (y - b)^{2}$ = $b^{2}$ is

The locus of the point of intersection of perpendicular tangents to the circles $x^{2} + y^{2} = a^{2} a n d x^{2} + y^{2} = b^{2} i s$

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

x^{2} - y^{2} = 0

x^{2} + y^{2} = a^{2}

{(x - c)}^{2} + y^{2} = a^{2}

x^{2} + {(y - b)}^{2} = a^{2}

\frac{x^{2}}{a^{2} + b^{2}} + \frac{y^{2}}{b^{2}} = 1, \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} + b^{2}} = 1

Join BYJU'S Learning Program