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.