The lines $2 \mathrm{x}-3 \mathrm{y}=5$ and $3 \mathrm{x}-4 \mathrm{y}=7$ are diameters of a circle having…
The lines $2 \mathrm{x}-3 \mathrm{y}=5$ and $3 \mathrm{x}-4 \mathrm{y}=7$ are diameters of a circle having area as 154 sq. units. Then the equation of the circle is
$x^2+y^2-2 x+2 y=62$
$x^2+y^2+2 x-2 y=62$
$x^2+y^2+2 x-2 y=47$
$x^2+y^2-2 x+2 y=47$
Solution
$\pi \mathrm{r}^2=154 \Rightarrow \mathrm{r}=7$
For centre on solving equation
$2 x-3 y=5 \quad \& \quad 3 x-4 y=7 \text { or } x=1, y=1 \text { centre }=(1,-1)$
Equation of circle, $(\mathrm{x}-1)^2+(\mathrm{y}+1)^2=7^2$
$x^2+y^2-2 x+2 y=47$