Standard form (x-h)² + (y-k)² = r² with worked example.

Find Equation of Circle Given Centre and Radius

Question

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

Solution — Step by Step

The equation of any circle with centre $(h, k)$ and radius $r$ is:

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

We use this form because every point $(x, y)$ on the circle is exactly $r$ units away from the centre — and this equation is just the distance formula squared.

From the problem: $h = 3$ , $k = -2$ , $r = 5$ .

Write these down before substituting — one sign error here ruins the whole answer.

Plug in the values:

(x - 3)^2 + (y - (-2))^2 = 5^2

(x - 3)^2 + (y + 2)^2 = 25

The equation of the circle is:

\boxed{(x-3)^2 + (y+2)^2 = 25}

This is the answer in standard form. NCERT sometimes asks you to expand it — we’ll do that in the Alternative Method below.

Why This Works

The standard form comes directly from the definition of a circle: the set of all points equidistant from a fixed centre. If $(x, y)$ is any point on the circle and $(h, k)$ is the centre, the distance between them equals $r$ .

The distance formula gives $\sqrt{(x-h)^2 + (y-k)^2} = r$ . We square both sides to get the clean standard form. No approximations, no tricks — just the Pythagoras theorem dressed up.

This is a Class 11 NCERT direct formula question. In JEE Main, this concept appears embedded in harder problems — finding the circle passing through three points, or finding tangent lengths — so getting the sign conventions right here pays off later.

Alternative Method — Expanded General Form

Sometimes a question or answer key gives the general form: $x^2 + y^2 + 2gx + 2fy + c = 0$ .

Expand our standard form:

(x-3)^2 + (y+2)^2 = 25

x^2 - 6x + 9 + y^2 + 4y + 4 = 25

x^2 + y^2 - 6x + 4y + 13 = 25

x^2 + y^2 - 6x + 4y - 12 = 0

Here $g = -3$ , $f = 2$ , $c = -12$ . You can verify: centre $= (-g, -f) = (3, -2)$ ✓ and radius $= \sqrt{g^2 + f^2 - c} = \sqrt{9 + 4 + 12} = \sqrt{25} = 5$ ✓

Always verify by recovering the centre and radius from your expanded form. Two seconds of checking saves marks in board exams.

Common Mistake

When the $k$ value is negative — like $k = -2$ here — students write $(y - 2)^2$ instead of $(y + 2)^2$ .

The formula has $(y - k)$ . With $k = -2$ : $(y - (-2)) = (y + 2)$ . The double negative flips to a plus. This is the single most common error on this type of problem — especially under exam pressure.

Question

Solution — Step by Step

Why This Works

Alternative Method — Expanded General Form

Common Mistake

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