conditions that the chord subtend a right angle

Question

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

can someone please help me out with the question, I simply can't find a way to approach it.

Answer 1

The equation of the circle is given as: $$x^2+y^2=a^2\tag 1 $$ and the point $P (x',y') $. The equation of the chord of contact of $P $ with respect to the equation of the circle will be: $$\frac {xx'}{a^2} + \frac {yy'}{a^2} = 0 \tag 2$$

To get the combined equation of the lines which join the origin, the points of intersection of $(1),(2) $; make $(1) $ homogenous with the help of $(2) $ giving us: $$x^2+y^2-a^2\left (\frac {xx'}{a^2}+\frac {yy'}{a^2}\right)=0$$

As the chord of contact of tangents are perpendicular, just equate the sum of coefficients of $x^2$ and $y^2$ to zero.

Answer 2

The centre, the two points of contact and the point $(x',y')$ form a square. So, $(x',y')$ is $\sqrt{2}a$ away from the centre. So $(x',y')$ lies on the circle $x^2+y^2=2a^2$.

Answer 3

A bit of detail:

Let tangents drawn from $P$ touch the circle in $A,A$'.

1)$\triangle OAP$ is a right triangle, $OA$ is perpendicular to tangent $AP.$

2)Angle subtended by chord is a right angle.

3) $OP$ is symmetry axis, and bisector of the subtended right angle, hence $\angle POA= \angle POA' = 45°$.

4) Hence $\angle OPA = 45°.$

5) $OP$ is the hypotenuse of the right , isosceles $\triangle OAP $ with legs of length $a$.

Pythagoras: $OP = √2 a.$

Hence:

The locus of points $P$ such that the chord of contact of tangents from $P$ subtends a right angle at the centre is a circle with radius $√2a$ about the origin.