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.