Question

The equation of the circle with centre $(–3,2)$ and radius $4\text{units}$ is

• A. $x^2+y^2+6x-4y-13=0$
• B. $x^2+y^2+6x-4y-3=0$
• C. $x^2+y^2+6x-4y-7=0$
• D. $x^2+y^2+6x-4y+3=0$

Answer 1

The correct option is B $x^2+y^2+6x-4y-3=0$

We know that, equation of the circle with centre $(h,k)$ and radius $r$ is $(x-h{)}^2+(y-k{)}^2=r^2$

Given, centre $:(–3,2)$ and radius $=4$
$\therefore (x-(-3){)}^2+(y-2{)}^2=4^2\Rightarrow (x+3{)}^2+(y-2{)}^2=16\Rightarrow x^2+y^2+6x-4y-3=0$