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Equation of a Circle Practice Problems

Fourteen hand-built problems on the equation of a circle, each fully worked. They cover reading the centre and radius off standard form, building the equation from a centre and a radius or a diameter, completing the square to convert from general form, testing whether a point is inside, and the two cases where the equation is not a circle at all.

  1. 1.

    What is the radius of \((x - 4)^2 + (y - 1)^2 = 9\)?

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    The right-hand side is \(r^2\), so \(r = \sqrt{9} = 3\).

  2. 2.

    What is the \(x\)-coordinate of the centre of \((x + 5)^2 + (y - 1)^2 = 49\)?

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    The equation subtracts the centre coordinates, so \((x + 5)^2\) means \(h = -5\). Rewriting it as \((x - (-5))^2\) makes the sign obvious.

  3. 3.

    What is the radius of \((x + 5)^2 + (y - 1)^2 = 49\)?

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    \(r = \sqrt{49} = 7\).

  4. 4.

    A circle has centre \((0,\ 0)\) and radius \(7\). Written as \(x^2 + y^2 = C\), what is \(C\)?

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    \(C = r^2 = 49\), giving \(x^2 + y^2 = 49\).

  5. 5.

    A circle has centre \((-3,\ 5)\) and radius \(2\). Written as \((x + 3)^2 + (y - 5)^2 = C\), what is \(C\)?

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    \(C = r^2 = 4\). Note the signs inside the brackets are the opposite of the centre's coordinates.

  6. 6.

    Complete the square on \(x^2 + y^2 - 8x + 2y + 8 = 0\). What is the radius?

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    \((x^2 - 8x + 16) + (y^2 + 2y + 1) = -8 + 16 + 1 = 9\), so \((x - 4)^2 + (y + 1)^2 = 9\) and \(r = 3\).

  7. 7.

    For \(x^2 + y^2 - 8x + 2y + 8 = 0\), what is the \(x\)-coordinate of the centre?

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    Half of \(-8\) is \(-4\), so the completed square is \((x - 4)^2\) and \(h = 4\).

  8. 8.

    Complete the square on \(x^2 + y^2 + 6x - 10y = 2\). What is the radius?

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    \((x + 3)^2 + (y - 5)^2 = 2 + 9 + 25 = 36\), so \(r = 6\). Whatever you add on the left must be added on the right too.

  9. 9.

    A circle has centre \((2,\ 3)\) and passes through \((5,\ 7)\). What is its radius?

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    The radius is the distance from the centre to any point on the circle: \(\sqrt{3^2 + 4^2} = \sqrt{25} = 5\).

  10. 10.

    A circle has a diameter from \((0,\ 2)\) to \((8,\ 8)\). What is its radius?

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    The diameter is \(\sqrt{8^2 + 6^2} = 10\), so \(r = 5\). The centre is the midpoint \((4,\ 5)\).

  11. 11.

    Is the point \((1,\ 1)\) inside, on, or outside \((x - 4)^2 + (y - 5)^2 = 25\)? Answer with one word.

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    Substitute: \((1-4)^2 + (1-5)^2 = 9 + 16 = 25\), exactly \(r^2\). The point lies on the circle.

  12. 12.

    What does the graph of \(x^2 + y^2 + 2x - 4y + 10 = 0\) look like? Answer "circle", "point", or "nothing".

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    \((x + 1)^2 + (y - 2)^2 = -10 + 1 + 4 = -5\). Two real squares cannot add to a negative, so no graph exists.

  13. 13.

    What does the graph of \(x^2 + y^2 - 4x + 6y + 13 = 0\) look like? Answer "circle", "point", or "nothing".

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    \((x - 2)^2 + (y + 3)^2 = -13 + 4 + 9 = 0\). The only way two squares sum to zero is if both are zero, so the graph is the single point \((2,\ -3)\).

  14. 14.

    Find the radius of \(2x^2 + 2y^2 - 12x + 8y = 6\).

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    Divide the whole equation by 2 first: \(x^2 + y^2 - 6x + 4y = 3\). Then \((x - 3)^2 + (y + 2)^2 = 3 + 9 + 4 = 16\), so \(r = 4\).
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Equation of a Circle Practice Problems

The Circles lesson shows where the equation comes from — it is the distance formula with the distance fixed — and this set is fourteen problems to practise on, each fully solved.

How to use this set

Most problems ask for a single value, so type just the number: a radius, a coordinate, or the constant in a given arrangement. The three classification problems want one word.

Work the completing-the-square problems on paper. The step people skip is adding the same amount to the right side, and it is much easier to spot on paper than in your head.

Common mistakes to watch for

Reporting as the radius. means .

Getting the centre’s signs backwards. means .

Adding only to the left when completing the square. That changes the radius.

Halving after squaring. The rule is half first, then square: for , half is , squared is .

Not dividing out a leading coefficient. must be divided by 2 first.

Where to go next

The conversion technique is the same one in Completing the Square. For the wider coordinate-geometry toolkit, see x- and y-Intercepts and Domain and Range.

Frequently Asked Questions

What is the standard equation of a circle?+

, where is the centre and is the radius. Centred at the origin it simplifies to .

Why are the signs inside the brackets flipped?+

The equation subtracts the centre's coordinates, so a centre of appears as . Read it back by negating what you see.

Is the right-hand side the radius?+

No, it is the radius squared. If the equation reads , the radius is . Forgetting the square root is the single most common slip here.

How do I complete the square for a circle?+

Group the -terms and -terms, then for each group add the square of half its linear coefficient — to **both** sides. For , half of is , and .

What if there is a number in front of x squared?+

Divide the entire equation by it first. A genuine circle has equal coefficients on and ; if they differ, the curve is an ellipse rather than a circle.

When is the equation not a circle?+

After completing the square, a right-hand side of zero gives a single point, and a negative right-hand side gives no graph at all. Both are legitimate answers, not mistakes.

More practice problems

Completing the Square Practice Problems x- and y-Intercepts Practice Problems Domain and Range Practice Problems