Question

In the equation $(x-3{)}^2$+$(y-2{)}^2$=$16$, the center of the circle is

• A. $(3,2)$
• B. $(-3,2)$
• C. $(3,-2)$
• D. $(2,3)$

Answer 1

The correct option is A $(3,2)$

$(x-3{)}^2+(y-2{)}^2=16---\left(i\right)$

General standard equation of circle$\Rightarrow x^2+y^2=r^2---\left(ii\right)$

Comparing $\left(i\right)$ and $\left(ii\right)$, $\left(ii\right)$ have centre at $(0,0)$. In $\left(i\right)$ only the origin is shifted. So, centre of $\left(i\right)$ $\Rightarrow (h,k):(3,2)$