Question

The locus of the image of the origin in line rotating about the point (1,1) is ?

Answer 1

Let P(x, y) be some point on the locus.
From the given information, we can see that the mid-point of O and P, $Q(\frac{x}{2}, \frac{y}{2})$, lies on the arbitrary line passing through O'(1, 1).
And lines OP & O'Q are perpendicular then the product of their slopes is -1.
Hence, $x(\frac{x}{2}-1)+y(\frac{y}{2}-1)=0$
$x(x-2)+y(y-2)=0$ → represents a Circle ( Centre=(1, 1) Radius=√2 )

Answer 2

Please see the diagram.

Let the point about which the line rotates  be C(1,1).  When the Line L rotates, it always passes through that with the fixed point being C.

Draw one line L through C(1,1) with a slope = m.

Image of O(0,0) is found by drawing a normal to L meeting L at Q  and extending to P, such that OQ = QP.

Since we are interested in locus of P.  Take its coordinates as P(x₁, y₁).   After finding a relation between x₁ and y₁, we will replace them by x and y.

Now  Q is the midpoint of OP.  Hence, Q = (x₁/2, y₁/2)

Now slope of L1 or  OP = (y₁-0)/(x₁-0) = -1/m , as it is perpendicular to L. 
       Slope of  QC = m = (1 - y₁/2) / (1 - x₁/2)

  Multiply slopes,    (1 - y₁/2) / (1 - x₁ /2 ) * (y₁ / x₁) = -1

           y₁ - y₁²/2 = - x₁ + x₁²/2
           x₁² + y₁² - 2x₁ - 2 y₁ = 0

             (x₁ - 1)² + (y₁ - 1)² = 2  = (√2)²

 Locus =  (x - 1)² + (y - 1)² = (√2)²

circle with C(1,1) as the center and  √2 as the radius.