Question

Consider a circle $C_1$ with equation $(x-1{)}^2+y^2=1$ and a shrinking circle $C_2$ with radius $r$ and centre at the origin. $P$ is the point $(0,r),Q$ is the point of intersection of the two circles(above x-axis) and $R$ is the point of intersection of the line $PQ$ with the x-axis. Find the x-coordinate of the point $R,$ as $C_2$ shrinks, that is $r\to 0$.

• A. 4
• B. 2
• C. 0
• D. 1

Answer 1

The correct option is A

4

Given, circle 1 : $C_1\equiv (x-1{)}^2+y^2=1$

circle 2 : $C_2\equiv x^2+y^2=r^2$

Finding the point of intersection of the two circles:

Putting $y^2=r^2-x^2$ in $C_1$, we get,

$(x-1{)}^2+r^2-x^2=1$

$\Rightarrow x=\frac{r^2}{2}$

$\Rightarrow y=\sqrt{r^2-\frac{r^4}{4}}$

We get $Q\equiv (\frac{r^2}{2},\sqrt{r^2-\frac{r^4}{4}})$

Given, $P\equiv (0,r)$

Let $R\equiv (x,0)$

Since, $P,Q,$ and $R$ are collinear points

Therefore, Slope of $PR=$ slope of $PQ$

$\Rightarrow \frac{r-0}{0-x}=\frac{r\sqrt{1-\frac{r^2}{4}}-r}{\frac{r^2}{2}-0}$

$\Rightarrow x={lim}_{r\to 0}\frac{r^2}{2(1-\sqrt{1-\frac{r^2}{4}})}$

$={lim}_{r\to 0}2\sqrt{4-r^2}$ [L'Hospital's rule]

Thus, as $r\to 0,x\to 4$