If $x^2=\lambda$, then find the value of $\lambda$

Question

A Circle $C_1$ is drawn having any point $P$ on $X$- axis as its centre and passing through the centre of the circle $C: x^2+y^2=1$. A common tangent to $C_1$, and $C$ intersects the circle at $Q$ and $R$ respectively. Then $Q(x,y)$ always satisfies $x^2 = \lambda $, find the value of $\lambda .$

Equation of $C_1$ will be $(x-h)^2+y^2=r^2$ and it passes through origin hence equation will be $(x-h)^2+y^2=h^2$. Now I write equation of tangent of slope $m$ for both circles but how do I get $x^2 = \lambda $ type equation?

Answer 1

The solution is quite straightforward if you use a different form of equation for a line: $\mathbf n\cdot\mathbf r=d$, where $\mathbf n$ is a fixed vector perpendicular to the line. One reason that this form is especially convenient here is that the distance of the line from the origin is $d/\|\mathbf n\|$.

Let $P=(h,0)$. A tangent to a circle is perpendicular to the radius at the point of tangency, so an equation of the tangent at $Q$ is $$\begin{align}(x_Q-h)x+y_Qy&=(x_Q-h)x_Q+y_Q^2\\&=(x_Q-h)^2+y_Q^2+h(x_Q-h)\\&=h^2+h(x_Q-h)\\&=hx_Q.\end{align}$$ Since this line is also tangent to $C$, its distance from the origin must be equal to the radius of $C$, which we can take to be an arbitrary value $r\gt0$ without complicating the solution. Thus,$${hx_Q\over\sqrt{(x_Q-h)^2+y_Q^2}}={h\over|h|}x_Q=r.$$ Note that we must have $|h|\ge\frac12r\gt0$ for there to be a common tangent line. Squaring both sides gives us $$x_Q^2=r^2.$$

Another way to appoach the problem is to express the slope of the tangent line in terms of $Q$, $R$ and $Q-R$, respectively, equate them all, and then try to eliminate the coordinates of $R$, but that felt like it would get messy quickly.

Answer 2

Hint: Let $A$ be the point on $C$ closest to $P$. Draw the perpendicular $OT$ from the center $O$ of the circle $C$ onto $PQ$. Show that $OPT$ and $QPA$ are congruent triangles via $OP=QP$, $\angle OPT=\angle QPA$, and $PT=PA$. This proves that $\angle QAP=\angle OTP$ is a right angle.

It is also possible to construct the point $P$, given $Q$ on the tangent to $C$ at $A$. Reflect $A$ about the line $OQ$ to get the point $R$. The perpendicular to $QR$ at $Q$ then meets the line $OA$ at $P$.