Question

Convert the equation \(r=2(h \cos \theta+k \sin \theta)\) to rectangular form and verify that it is the equation of a circle. Find the radius and the rectangular coordinates of the center of the circle.

Short answer

The circle's center is at the point (h, k) and the radius of the circle is \(\sqrt{h^2 + k^2}\).

Step-by-step solution

Conversion from Polar to Rectangular

The equation \(r=2(h \cos \theta+k \sin \theta)\) is given in polar form. To convert this into rectangular coordinates, we can substitute \(r = \sqrt{x^2 + y^2}\), \(\cos \theta = x / r\), and \(\sin \theta = y / r\) into the given equation. This gives us \(\sqrt{x^2 + y^2} = 2(h (x / \sqrt{x^2 + y^2}) + k (y / \sqrt{x^2 + y^2}))\). This simplifies to \(\sqrt{x^2 + y^2} = 2hx/r + 2ky/r\). Multiply by \(\sqrt{x^2 + y^2}\) and simplify to get \(x^2 + y^2 = 2hx + 2ky\).

Confirm equation as Circle

The general equation for a circle is \((x-a)^2 + (y-b)^2 = r^2\). To rewrite the equation \(x^2 + y^2 = 2hx + 2ky\) in that form, move the terms around to get \((x-h)^2 + (y-k)^2 = h^2 + k^2\). This is equivalent to the general equation of circle which confirms our circle.

Find the Center and Radius

Comparing \((x-h)^2 + (y-k)^2 = h^2 + k^2\) to the general circle equation \((x-a)^2 + (y-b)^2 = r^2\), we can identify \(a = h\), \(b = k\), and \(r = \sqrt{h^2 + k^2}\) which gives us the center and radius of the circle.