Q) A circle $C_{1}$ is drawn having 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 $C$ touches the circles at Q and R
respectively . Then $Q(x,y)$ always satisfies $x^{2}=\lambda $ , then
find $\lambda$ ?
Attempt
Let $(p,0) $ be the centre of $C_1$ then we have $C_1 = x^2 +y^2 -2px=0$.
Let $R=(x_1 ,y_1 )$ and $Q=(x_2 , y_2 )$
Then I wrote the equation of tangents of both circles and equated them and got
$\frac{1}{p}=x_1 + x_2$
How do I proceed? Hints?
Best Answer
$\boldsymbol{r\lt1}$
Using similar triangles, we get the $x$-coordinate of $Q$ to be $r+(1-r)=1$.
The sum of the orange and lavender segments times $r$ should be the length of the lavender segment; that is, $$ \frac{\color{#C00000}{r}}{\color{#00A000}{1}}(\color{#FF8000}{r}+\color{#8080FF}{x})=\color{#8080FF}{x} $$ solving gives $$ x=\frac{r^2}{1-r} $$
$\boldsymbol{r\gt1}$
Using similar triangles, we get the $x$-coordinate of $Q$ to be $r-(r-1)=1$.
The sum of the lavender and orange segments should be $r$ times the length of the lavender segment; that is, $$ \color{#8080FF}{x}+\color{#FF8000}{r}=\frac{\color{#C00000}{r}}{\color{#00A000}{1}}\color{#8080FF}{x} $$ solving gives $$ x=\frac{r}{r-1} $$
If the center of $C_1$ is on the left of $(0,0)$, then the $x$-coordinate of $Q$ will be $-1$.