The equation of a circle passing through origin and making $\mathrm{x}$ -intercept 3 and $\mathrm{y}$ -…
The equation of a circle passing through origin and making $\mathrm{x}$ -intercept 3 and $\mathrm{y}$ -
intercept $-5$ is
$x^{2}+y^{2}+3 x+5 y=0$
$x^{2}+y^{2}+3 x-5 y=0$
$x^{2}+y^{2}-3 x+5 y=0$
$x^{2}+y^{2}-3 x-5 y=0$
Solution
$\because A(3,0), B(0,-5)$ be the co-ordinate of ends of diameter AB. By diameter form, equation of circle is
$\begin{array}{l}
(x-3)(x-0)+(y-0)(y+5)=0 \\
x^{2}-3 x+y^{2}+5 y=0 \Rightarrow x^{2}+y^{2}-3 x+5 y=0
\end{array}$
