Circle equations are “geometry in coordinates.” Once you can move between standard form and general form, you can find the center and radius instantly, graph circles cleanly, and solve intersection questions with lines.
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If the equation is in this form, the center is (h, k) and the radius is r. The biggest “gotcha” is sign: if you see (x + 3)^2, then h = −3.
Problem: Find the center and radius of (x − 2)^2 + (y + 5)^2 = 49.
Notice (y + 5)^2 means y is shifted down 5, so k = −5.
A “general form” circle equation might look like x^2 + y^2 + Dx + Ey + F = 0. To convert it to standard form, you complete the square for x and for y.
Problem: Convert x^2 + y^2 − 6x + 4y − 12 = 0 to standard form and find the center/radius.
Group and move the constant:
Complete the square:
Add 9 and 4 to both sides to keep balance:
Center = (3, −2) and radius = 5.
To graph a circle from standard form, plot the center, then go r units up/down/left/right to mark four key points. Sketch a smooth curve through them.
Problem: Write the equation of the circle with center (−1, 3) that passes through the point (2, 7).
Use standard form. The only missing piece is r^2. Since radius is the distance from the center to the point, compute distance squared to avoid square roots:
So the equation is (x + 1)^2 + (y − 3)^2 = 25.
Problem: A circle has a diameter with endpoints A(0, 2) and B(6, 10). Find the circle equation.
The center is the midpoint of the diameter. Then use half the diameter length as the radius.
Equation: (x − 3)^2 + (y − 6)^2 = 25.
A common “circle equation” task is finding where a line intersects a circle. The method is consistent: substitute the line into the circle, solve the resulting quadratic, then interpret the result.
Tip: if you’re asked for “how many intersection points,” you can often decide by whether the quadratic has 2, 1, or 0 real roots.
If you’re converting from general form often, it helps to follow the exact same micro-steps every time. The goal is to create perfect squares in x and y.
This “half then square” rule is the part students forget most. If you keep it consistent, circle conversions become routine.
Problem: Decide whether (x − 1)^2 + (y + 2)^2 = −9 represents a real circle.
In standard form, the right side is r^2. But a real radius squared cannot be negative.
Interpretation: there is no point (x,y) whose squared distances add to a negative number, so the graph is empty.
(x + 4)^2 means h = −4. Many answers are off by just a sign.
When completing the square, whatever you add inside must be balanced by adding the same amount to the other side.
If you end up with r^2 < 0, there is no real circle.
Circles use x^2 and y^2 together. If only one variable is squared, it’s not a circle.
For a quick reference on standard form and completing-the-square conversion, the analytic geometry notes on OpenStax conic sectionsinclude circle form and the same conversion technique.
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