Find the equation of the circle whose centre is in the 2nd quadrant, touches soth the xaxis and yaxis as tangents and whose radius is 4 units in length.

The equation of a circle is given by \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius. Since the radius is 4 units and the circle touches both the x-axis and y-axis, the center must be located at \((-4, 4)\) in the 2nd quadrant (since it is above the x-axis and to the left of the y-axis). Plugging these values into the circle's equation gives you: \[ (x + 4)^2 + (y - 4)^2 = 16. \] Now, you have the equation of the circle: \[ (x + 4)^2 + (y - 4)^2 = 16. \] This captures both the center at \((-4, 4)\) and the radius of 4.