If the lines x + 2y - 5 = 0 and 3x - y - 1 = | vedclass

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(B) The center of the circle is the point of intersection of its diameters.Given equations of diameters:$x + 2y - 5 = 0$ ... $(i)$$3x - y - 1 = 0$ ...

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$x^2 + y^2 - 2x - 4y - 20 = 0$

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(B) The center of the circle is the point of intersection of its diameters.Given equations of diameters:$x + 2y - 5 = 0$ ... $(i)$$3x - y - 1 = 0$ ... $(ii)$Multiplying $(ii)$ by $2$,we get $6x - 2y - 2 = 0$ ... $(iii)$Adding $(i)$ and $(iii)$:$(x + 2y - 5) + (6x - 2y - 2) = 0$$7x - 7 = 0 \Rightarrow x = 1$Substituting $x = 1$ in $(i)$:$1 + 2y - 5 = 0$ $\Rightarrow 2y = 4$ $\Rightarrow y = 2$So,the center $(h, k) = (1, 2)$ and the radius $r = 5$.The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.$(x - 1)^2 + (y - 2)^2 = 5^2$$x^2 - 2x + 1 + y^2 - 4y + 4 = 25$$x^2 + y^2 - 2x - 4y + 5 - 25 = 0$$x^2 + y^2 - 2x - 4y - 20 = 0$

If the lines $x + 2y - 5 = 0$ and $3x - y - 1 = 0$ denote two diameters of a circle of radius $5 \text{ units}$,then the equation of the circle is

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