If the lines $2 x-3 y=5$ and $3 x-4 y=7$ are two diameters of a circle of radius 7 , then the equation of…
If the lines $2 x-3 y=5$ and $3 x-4 y=7$ are two diameters of a circle of radius 7 , then the equation of the circle is
$x^2+y^2+2 x-4 y-47=0$
$x^2+y^2=49$
$x^2+y^2-2 x+2 y-47=0$
$x^2+y^2=17$
Solution
Since, the lines $2 x-3 y=5$ and $3 x-4 y=7$ are the diameters of a circle. Therefore, the point of intersection is the centre of the circle. On solving the given equations, we get $x=1$ and $y=-1$ ie, the centre of the circle.
$\therefore$ Required equation of circle is
$
\begin{aligned}
& (x-1)^2+(y+1)^2 & =7^2 \\
\Rightarrow & x^2+y^2-2 x+2 y+2 & =49 \\
\Rightarrow & x^2+y^2-2 x+2 y-47 & =0
\end{aligned}
$