A circle $C_{1}$ is drawn having any point $P$ on axis as its centre and passing through the centre of the circle $C : x^{2} + y^{2} = a$ . A common tangent to $C_{1}$ and C intersects the circles at $Q$ and $R$ respectively . Then , $Q (x, y)$ always satisfies $X^{2} = λ$ .

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Solution

Let the point be $P (h, 0)$ .
Equation of circle is $(x - h)^{2} + y^{2} = r^{2}$
It passes through centre of $x^{2} + y^{2} = 1$ which is $(0, 0)$ , hence $(0 - h)^{2} + 0 = r^{2}$
$\Rightarrow h^{2} = r^{2}$
Hence equation will be
$(x - h)^{2} + y^{2} = h^{2}$ .........(i)
A tangent to a circle is perpendicular to the radius at the point of tangency, so an equation of the tangent at Q is
$(x Q - h) + y Q y = (x Q - h) x Q + y^{2} Q$
$= x Q^{2} - h x Q + y^{2} Q$
$= x Q^{2} + h^{2} - 2 h x Q + h x Q - h^{2} + y^{2} Q$
$= (x Q - h)^{2} + y Q^{2} + h (x Q - h)$
$= h^{2} + h x Q - h^{2}$ (Using (i))
$(x Q - h) x + y Q y = h x Q$
$\Rightarrow (x Q - h) x + y Q y - h x Q = 0$
Since this line is also tangent to C, its distance from the origin must be equal to the radius of C.
$\frac{h x Q}{\sqrt{x Q - h)^{2} + y Q^{2}}} = 1$ (Radius of $x^{2} + y^{2} = 1$ is $1$ unit)
$\Rightarrow \frac{h x Q}{\sqrt{h^{2}}} = 1$
Square both sides
$\frac{h^{2} x Q^{2}}{h^{2}} = 1$
$\Rightarrow x Q^{2} = 1$
$\Rightarrow λ = 1$ .

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