The equation of circle with centre at $(2,-3)$ and the circumference $10 \pi$ units is
The equation of circle with centre at $(2,-3)$ and the circumference $10 \pi$ units is
$x^2+y^2-4 x+6 y-12=0$
$x^2+y^2-4 x-6 y-12=0$
$x^2+y^2+4 x+6 y+12=0$
$x^2+y^2-4 x+6 y+12=0$
Solution
We have $2 \pi r=10 \pi \Rightarrow r=5$ and centre of circle is $(2,-3)$ Hence equation of circle is $(x-2)^2+(y+3)^2=(5)^2$ i.e. $x^2+y^2-4 x+6 y-12=0$