Question

Find the equation of the circle which passes through the origin and cuts off intercepts $a$ and $b$ respectively from $x$ and $y$-axis.

Answer 1

Since the circle is passes through origin and cut intercept $a$ and $b$ on $x$-axis and $y$-axis
$\Rightarrow$ Circle passes through the the points $A(a,0)$ and $B(0,b)$.

Since, $x$-intercept is $a$ and $y-$ intercept is $b$and the circle is passing through origin.

i.e., $AB$ is the diameter of circle.

If $(x_1,y_1)$ and $(x_2,y_2)$ are end points of diameter of the circle then, equation of the circle is,

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

Here, $A(a,0)$ and $B(0,b)$ are the end points of diameter of the circle.

So, the equation of the circle is,
$(x-a)(x-0)+(y-0)(y-b)=0$

$\Rightarrow (x-a)x+(y-b)y=0$

$\Rightarrow x^2+y^2-ax-by=0$--(1)

Since, the intercepts $a$ and $b$ may negative.

$\Rightarrow$ Required equation of the circle is,

$x^2+y^2\pm ax\pm by=0$