Let P(2,1) be a point in circle C with center at (0,0) and a radius of 3. For every point X on circle C, a perpendicular straight line is drawn through the midpoint of XP. The envelope of the family of all these straight lines is an ellipse. Find the equation of this ellipse.

The line A1: y = (1/2)x is a line of symmetry of the figure, so one of the ellipse's axis will lay on that line.

y = (1/2)x intersects the circle at X1=(6/sqrt[5], 3/sqrt[5]) and X2=(-6/sqrt[5], -3/sqrt[5]).

The two lines derived from X1 and X2 per the construction are L1: y = -2x + (5/2 + 3sqrt[5]/2) and L2: y = -2x + (5/2 - 3sqrt[5]/2).

The other axis of the ellipse is halfway between these two lines and then is A2: y = -2x + 5/2.

The line y = -2x + 5 is perpendictular to y = (1/2)x at the point (2,1).

y = -2x + 5 intersects the circle at X3=(2+2/sqrt[5], 1-4/sqrt[5]) and X4=(2-2/sqrt[5], 1+4/sqrt[5]).

The two lines derived from X3 and X4 per the construction are L3: y = (1/2)x - sqrt[5]/2 and L4: y = (1/2)x + sqrt[5]/2.

The ellipse then has axis on lines A1 and A2 and the bounding box around the ellipse consists of tangent lines L1, L2, L3, and L4.

Writing each of these lines in the form Ax+By+C=0 with A^2+B^2=1:

A1: (1/sqrt[5])x - (2/sqrt[5])y = 0

A2: (2/sqrt[5])x + (1/sqrt[5])y - sqrt[5]/2 = 0

L1: (2/sqrt[5])x + (1/sqrt[5])y - sqrt[5]/2 - 3/2 = 0

L2: (2/sqrt[5])x + (1/sqrt[5])y - sqrt[5]/2 + 3/2 = 0

L3: (1/sqrt[5])x - (2/sqrt[5])y - 1 = 0

L4: (1/sqrt[5])x - (2/sqrt[5])y + 1 = 0

From L1 and L2 the length of the axis on A1 is 3; similarly from L3 and L4 the length of the axis on A2 is 2.

Then the equation of the ellipse can be formed using these two values and the expressions for A1 and A2:

[(1/sqrt[5])x - (2/sqrt[5])y]^2 / (2/2)^2 + [(2/sqrt[5])x + (1/sqrt[5])y - sqrt[5]/2]^2 / (3/2)^2 = 1.

Expanding and simplifying eventually yields the equation 5x^2 - 4xy + 8y^2 - 8x - 4y - 4 = 0.