A circle has radius $3$ units and its centre lies on the line $y=x-1$. Then the equation of this circle if it passes through the point $(7, 3)$, is?

The radius of circle $x^2+y^2-6x-8y=0$

The equation of the circle having centre $(1,\ -2)$ ;and passing through the point of intersection of the lines $3x+y=14$ and $2x+5y=18$ is

$\mathrm{A}(1,1)$ বিন্দুটি $\mathrm{x}^{2}+\mathrm{y}^{2}+4 \mathrm{x}+6 \mathrm{y}-12=0$ বৃত্তের উপর অবস্থিত। রেখাত্রয়ের সমীকরণ $\mathrm{x}=0, \mathrm{y}=0, \mathrm{x}=\mathrm{a}$.