**Equation of a Circle**

The equation of the circle with center C (h, k) and radius r is (x – h)² + (y – k)² = r².

**Proof: **Above image represents P (x₁, y₁) is locus of the point, r is radius and C (h, k) is center.

The distance between the CP is r.

let P (x₁, y₁) be a point P lies in the circle **⇒** PC = r

**⇒** \(\sqrt{{{({{x}_{1}}-h)}^{2}}+{{({{y}_{1}}-k)}^{2}}}=r\).

**⇒** (x₁ – h) ² + (y₁ – k) ² = r²

The locus of P is (x – h)² + (y – k)² = r²

**Note:**

i) The equation of a circle of the form x² + y² + 2gx + 2fy + c =0

ii) The equation of a circle with center origin (0, 0) and radius r is

(x – 0)² + (y – 0)² = r²

x² + y² = r²

**Example 1: **Find the equation of the circle with center (1, 4) and radius 5.

**Solution:** Given that,

C (h, k) = (1, 4), P (x, y) and r = 5

The locus of P is (x-h) ² + (y-k) ² = r²

(x-1) ² + (y-4) ² = 5²

x² – 2x + 1 + y² – 8y + 16 = 25

x² + y² – 2x – 8y – 8 = 0

Required equation of the circle is x² + y² – 2x – 8y – 8 = 0

**Example 2: **Find the equation of the circle with center origin and radius 9.

**Solution: **Given that,

C (h, k) = (0, 0), P (x, y) and r = 9

The locus of P is (x-0) ² + (y-0) ² = r²

x² + y² = 9²

x² + y² – 81 = 0.