If the line $x+2 y-5=0$ and $3 x-y-1=0$ denote two diameters of a circle of radius 5 units, then the…
If the line $x+2 y-5=0$ and $3 x-y-1=0$ denote two diameters of a circle of radius 5 units, then the equation of the circle is
$x^2+y^2-2 x+4 y-20=0$
$x^2+y^2-2 x-4 y-20=0$
$x^2+y^2+2 x-4 y+20=0$
$x^2+y^2+2 x+4 y+20=0$
Solution
Equation of diameters of the circle,
x + 2y − 5 = 0 … (i)
3x − y − 1 = 0 … (ii)
Since, intersection of lines (i) and (ii) is the centre
of the circle.
On solving Eqs. (i) and (ii), we get x = 1 and y = 2
$\therefore$ Centre $=(1,2)$ and radius $=5$
$\therefore$ Equation of required circle
$
\begin{gathered}
(x-1)^2+(y-2)^2=(5)^2 \\
\Rightarrow x^2+y^2-2 x-4 y-20=0
\end{gathered}
$