Q.Find the equation of the circle with centre $(2,-3)$ and radius $5$.

With the centre and radius given directly, substituting into $(x-h_0)^2+(y-k_0)^2=r^2$ and expanding produces the equation in general form.

Substituting centre $(2,-3)$ and radius $5$ into $(x-h_0)^2+(y-k_0)^2=r^2$:

Expanding: $x^2-4x+4+y^2+6y+9=25$, so:

$x^2+y^2-4x+6y+13=25 \quad \Rightarrow \quad x^2+y^2-4x+6y-12=0$ …

Writing $(y-(-3))^2$ as $(y-3)^2$ instead of $(y+3)^2$ is the most common slip here — always carefully rewrite a double-negative before squaring. A second common error is arithmet …