What is the polar equation of a circle of radius √(a²+b²) centered at (a, b)?

Verified step by step guidance

1. Recall that the Cartesian equation of a circle centered at \((a, b)\) with radius \(r\) is given by \((x - a)^2 + (y - b)^2 = r^2\). In this problem, the radius is \(\sqrt{a^2 + b^2}\), so the equation becomes \((x - a)^2 + (y - b)^2 = a^2 + b^2\).
2. Express the Cartesian coordinates \(x\) and \(y\) in terms of polar coordinates \(r\) and \(\theta\) using the relations \(x = r \cos \theta\) and \(y = r \sin \theta\). Substitute these into the circle equation to get \((r \cos \theta - a)^2 + (r \sin \theta - b)^2 = a^2 + b^2\).
3. Expand the squares on the left-hand side: \((r \cos \theta)^2 - 2 a r \cos \theta + a^2 + (r \sin \theta)^2 - 2 b r \sin \theta + b^2 = a^2 + b^2\).
4. Combine like terms and simplify. Notice that \((r \cos \theta)^2 + (r \sin \theta)^2 = r^2\), and \(a^2 + b^2\) appears on both sides, so they cancel out. This leaves the equation \(r^2 - 2 a r \cos \theta - 2 b r \sin \theta = 0\).
5. Factor out \(r\) from the terms involving it: \(r^2 = 2 r (a \cos \theta + b \sin \theta)\). Since \(r \neq 0\), divide both sides by \(r\) to get the polar equation of the circle: \boxed{r = 2 (a \cos \theta + b \sin \theta)}.

Key Concepts

Polar Coordinates

Polar coordinates represent points in the plane using a distance from the origin (r) and an angle (θ) from the positive x-axis. Unlike Cartesian coordinates (x, y), polar coordinates express location based on radius and direction, which is essential for converting equations of curves into polar form.

Equation of a Circle in Cartesian Coordinates

A circle centered at (a, b) with radius R has the Cartesian equation (x - a)² + (y - b)² = R². Understanding this standard form is crucial before converting it into polar coordinates, as it provides the geometric definition and parameters of the circle.

Conversion Between Cartesian and Polar Coordinates

To convert Cartesian equations to polar form, use x = r cos θ and y = r sin θ. Substituting these into the Cartesian equation allows rewriting the curve in terms of r and θ, enabling the derivation of the polar equation of the circle.