Question

Find the equations of the circles which pass through the origin and cut off equal chords of \[\sqrt{2}\] units from the lines y = x and y = − x.

Answer 1

Suppose

\[a = \sqrt{2}\] From the figure, we see that there will be four circles that pass through the origin and cut off equal chords of length a from the straight lines \[y = \pm x\]

AB, BC, CD and DA are the diameters of the four circles.

Also,

\[C_1 A = \frac{a}{\sqrt{2}} = O C_1\] Thus, the coordinates of A are  \[\left( \frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}} \right)\]

In the same way, we can find the coordinates of B, C and D as

\[\left( \frac{- a}{\sqrt{2}}, \frac{a}{\sqrt{2}} \right),\] \[\left( \frac{- a}{\sqrt{2}}, \frac{- a}{\sqrt{2}} \right)\] and

\[\left( \frac{a}{\sqrt{2}}, \frac{- a}{\sqrt{2}} \right)\], respectively.
The equation of the circle with AD as the diameter is

\[\left( x - \frac{a}{\sqrt{2}} \right)\left( x - \frac{a}{\sqrt{2}} \right) + \left( y - \frac{a}{\sqrt{2}} \right)\left( y + \frac{a}{\sqrt{2}} \right) = 0\], which can be rewritten as

\[x^2 + y^2 - \sqrt{2}ax = 0\] , i.e. \[x^2 + y^2 - 2x = 0\]

Similarly, the equations of the circles with BC, CD and AB as the diameters are \[x^2 + y^2 + 2x = 0\]

\[x^2 + y^2 + 2y = 0\]  and  \[x^2 + y^2 - 2y = 0\], respectively.