Find the circle equation constants from points on the graph

Key concept: When a circle is written as $x^2+y^2=ax+by+c$, the clean way to recover the constants is to substitute coordinates from points that lie on the circle. That turns the picture into a small linear system.

Step by step

Step 1: Substitute the known points

For a point $(x,y)$ on the circle $x^2+y^2=ax+by+c,$ so every visible point gives one linear equation in $a,b,c$.

Using the three labeled points shown in the working:

• $(-3,3)$ gives $18=-3a-3b+c$
• $(3,4)$ gives $25=3a+4b+c$
• $(-1.5,2)$ gives $6.25=-1.5a+2b+c$

Step 2: Eliminate $c$

Subtract the first equation from the second: $7=6a+7b$

Subtract the first equation from the third: $-11.75=1.5a+5b$

Step 3: Solve the linear system

From $6a+7b=7$ and $1.5a+5b=-11.75,$ you solve for $a$ and $b$, then back-substitute to get $c$.

So the key method is: pick three points from the graph, build three equations, and solve the system.

If your diagram labels are meant to be different points, the final numbers will change, but the setup stays the same.

Pitfall alert

A common mistake is to plug in the center of the circle, or to read a label from the sketch as if it were a point on the curve when it is only an annotation. Another easy slip is mixing up $x^2+y^2=ax+by+c$ with the standard center-radius form. Here, do not expand into center form unless the question asks for it; just use the given points directly.

Try different conditions

If one of the marked points changes, the algebra is still identical: each new point gives a new equation of the form $x^2+y^2=ax+by+c$. If you only know two points, you cannot determine all three constants uniquely; you need a third independent point. If the circle is given in center form instead, you would first expand it before matching coefficients.

Further reading

circle equation, substitution method, linear system