Hint: First we will calculate the equation of tangent from the given point, then obtain the value of k and then apply the condition that the chord will only subtend 90 degrees if the chord itself is the diameter. Then after solving it further we got a condition we need to find out.

Complete step-by-step solution:

The diagram for the question is given below:

The equation of circle given is ${x^2} + {y^2} = {a^2}$

Let the point of contact of the tangent be (h, k).

The tangent at this point in the circle is given by:

$hx + hy = {a^2}$

Since this tangent passes through the point (x’, y’). We can write:

$hx' + hy' = {a^2}$

Also, the equation of the circle given is ${x^2} + {y^2} = {a^2}$and point (h, k) present on the circle. So, we have:

${h^2} + {k^2} = {a^2}$

Using these two equations we have:

${h^2} + {\left( {\dfrac{{{a^2} - hx'}}{{y'}}} \right)^2} = {a^2}$

On solving it further we get,

$ \Rightarrow y{'^2}{h^2} + {a^4} - 2{a^2}hx' + {h^2}x{'^2} = {a^2}y{'^2}$

$ \Rightarrow {h^2}(y{'^2} + x{'^2}) - 2{a^2}hx' + {a^4} - {a^2}y{'^2} = 0$

We know that the chord will only subtend 90 degree if the chord itself is the diameter.

We know the equation of circle having its diametric ends at the two points of contacts can be written as,

$(x - {h_1})(x - {h_2}) + (y - {k_1})(y - {k_2}) = 0$

$ \Rightarrow {x^2} - x({h_1} + {h_2}) + {h_1}{h_2} + {y^2} - y({k_1} + {k_2}) + {k_1}{k_2} = 0$

Since the chord of contact is the diameter, the circle passes through the origin .

So, we have:

${h_1}{h_2} + {k_1}{k_2} = 0$

On putting the value of${k_1}{k_2}$ in the above equation we get,

${h_1}{h_2} + \dfrac{{({a^2} - {h_1}x')({a^2} - {h_2}x')}}{{y{'^2}}} = 0$

Then solving it further we get,

$ \Rightarrow {h_1}{h_2}y{'^2} + {a^4} - {a^2}x'({h_1} + {h_2}) + {h_1}{h_2}x{'^2} = 0$

\[ \Rightarrow \left( {x{'^2} + y{'^2}} \right) \times \dfrac{{{a^4} - {a^2}y{'^2}}}{{x{'^2} + y{'^2}}} + {a^4} - {a^2}x{'^2} \times \left( {\dfrac{{2{a^2}x'}}{{x{'^2} + y{'^2}}}} \right) = 0\]

$ \Rightarrow {a^4} - {a^2}y{'^2} + {a^4} - \left( {\dfrac{{2{a^4}x{'^2}}}{{x{'^2} + y{'^2}}}} \right) = 0$

\[ \Rightarrow (2{a^4} - {a^2}y{'^2})(x{'^2} + y{'^2}) - 2{a^4}x{'^2} = 0\]

On further solving we have:

\[ \Rightarrow 2{a^4}y{'^2} - {a^2}x{'^2}y{'^2} - {a^4}y{'^2} = 0\]

\[ \Rightarrow 2{a^2} - x{'^2} - y{'^2} = 0\]

After transposing we get,

\[ \Rightarrow x{'^2} + y{'^2} = 2{a^2}\]

This is the required condition we need to find.

Note: In this type of question, you should know to write the equation of the tangent at a point on the circle. In this particular question, the four points i.e. the two points of contact of tangents, the point (x’, y’), and the center will form a square with side length ‘a’.