Question

Find the condition that the chord of contact of tangents from the point $(x^{\prime },y^{\prime })$ to the circle $x^2+y^2=a^2$ should subtend a right angle at the centre.

Answer 1

Let the point of contact of the tangent be $(h,k)$.
The tangent at this point to the given circle is given by $hx+ky=a^2$
Since this passes through the point $(x^{\prime },y^{\prime })$, we have $h{x}^{\prime }+k{y}^{\prime }=a^2$
Also, $h^2+k^2=a^2$
Using these two equations, we have
$h^2+{\left(\frac{a^2-h{x}^{\prime }}{y^{\prime }}\right)}^2=a^2$
$\Rightarrow y^{\prime 2}{h}^2+a^4-2a^{2}h{x}^{\prime }+h^2{x}^{\prime 2}=a^2{y}^{\prime 2}$
$\Rightarrow h^2(x^{\prime 2}+y^{\prime 2})-h\left(2a^2{x}^{\prime }\right)+a^4-a^2{y}^{\prime 2}=0$
Now, the equation of the circle having its diametric ends as the two points of contact can be written as $(x-h_1)(x-h_2)+(y-k_1)(y-k_2)=0$
i.e. $x^2-x(h_1+h_2)+h_1{h}_2+y^2-y(k_1+k_2)+k_1{k}_2=0$
Since the chord of contact is the diameter, the circle passes through the origin.
$\Rightarrow h_1{h}_2+k_1{k}_2=0$
i.e. $h_1{h}_2+\frac{(a^2-h_1{x}^{\prime })(a^2-h_2{x}^{\prime })}{y^{\prime 2}}=0$
$\Rightarrow h_1{h}_2{y}^{\prime 2}+a^4-a^2{x}^{\prime }(h_1+h_2)+h_1{h}_2{x}^{\prime 2}=0$
$\Rightarrow (x^{\prime 2}+y^{\prime 2})\times \frac{a^4-a^2{y}^{\prime 2}}{x^{\prime 2}+y^{\prime 2}}+a^4-a^2{x}^{\prime }\times \left(\frac{2a^2{x}^{\prime }}{x^{\prime 2}+y^{\prime 2}}\right)=0$
$\Rightarrow a^4-a^2{y}^{\prime 2}+a^4-\frac{2a^4{x}^{\prime 2}}{x^{\prime 2}+y^{\prime 2}}=0$
$\Rightarrow (2a^4-a^2{y}^{\prime 2})(x^{\prime 2}+y^{\prime 2})-2a^4{x}^{\prime 2}=0$
$\Rightarrow 2a^4{y}^{\prime 2}-a^2{x}^{\prime 2}{y}^{\prime 2}-a^2{y}^{\prime 4}=0$
i.e. $2a^2-x^{\prime 2}-y^{\prime 2}=0$
i.e. $x^{\prime 2}+y^{\prime 2}=2a^2$