How to find the constants in a circle equation from its centre and radius

Expert intro: Circle equations often look messy in expanded form, but once you spot the centre and radius, the constants fall into place quickly.

Detailed walkthrough

Start from the circle in standard form:

$(x-h)^2+(y-k)^2=r^2.$

From the given solution, the circle is

$(x-3)^2+(y+2)^2=6^2.$

So the centre is $(3,-2)$ and the radius is $6$.

Expand the equation

$(x-3)^2=x^2-6x+9,$

$(y+2)^2=y^2+4y+4.$

Add them:

$x^2-6x+9+y^2+4y+4=36.$

Rearrange:

$x^2+y^2=6x-4y+23.$

Compare this with

$x^2+y^2=ax+by+c.$

Therefore,

$\boxed{a=6,\quad b=-4,\quad c=23}.$

Quick check

• The coefficient of $x$ is positive because the centre’s $x$-coordinate is 3.
• The coefficient of $y$ is negative because the centre’s $y$-coordinate is $-2$.
• The constant comes from $9+4$ moved across from $36$, giving $23$.

💡 Pitfall guide

Students often flip the sign when expanding $(y+2)^2$ or forget to move the constant term after expansion. Another easy trap is mixing up the centre coordinates with the signs in the expanded equation: centre $(h,k)$ gives coefficients $-2h$ and $-2k$ after expansion.

🔄 Real-world variant

If the centre were $(h,k)$ and radius $r$, then

$(x-h)^2+(y-k)^2=r^2$

expands to

$x^2+y^2=2hx+2ky+(r^2-h^2-k^2).$

That template lets you identify the constants almost immediately.

🔍 Related terms

standard form of a circle, centre and radius, circle expansion