What changes can be made to the question or the solution so that the answer is $2$?

Question

If tangent $PR$ of length $4$ unit is drawn at end point of diameter PQ of the circle $S_i:x^2+y^2=4$. If line $QR$ meets circle at point $S$. Another circle $S_2$ is drawn taking $SQ$ as diameter. If angle between $S_1$ and $S_2$ is $\alpha$ then find $|\tan\alpha|$.

Solution: I found its solution. But I think there is some discrepancy in the solution or question. Not able to figure that out.

$PS=x, SQ=y$

$x^2+y^2=4$

$\implies \tan\theta=\frac{PR}{PQ}=2=\frac xy$

$\implies\cos\theta=\frac1{√5}=\frac{\frac y2}1=y=\frac2{√5}$

$\implies x=\frac4{√5}$

$\implies C_1C_2=\frac12PS=\frac2{√5}$

$|\cos\alpha|=|\frac{r_1^2+r_2^2-(C_1C_2)^2}{2nr_2}|=|\frac{1+\frac15-\frac45}{2(1)(\frac1{√5})}|$

$|\cos\alpha|=\frac1{√5}\implies|\tan\alpha|=2$

Answer 1

Let $r_i$ be the radius of $S_i$ for $i=1,2$.

First of all, I think that "$S_{\color{red}{i}}:x^2+y^2=4$" should be "$S_{\color{red}1}:x^2+y^2=4$".

Then, I think that $\alpha$ is equal to either $\angle{TSP}$ or $\pi−\angle{TSP}$, and that $\angle{TSP}=θ$ because the angle between a chord and the tangent line at one of its intersection points equals half of the central angle subtended by the chord (see here) where $ST$ is tangent to $S_1$ at $S$. Therefore, we get $|\tan\alpha|=\tan\theta$.

So, if $r_1=2$, then $|\tan\alpha|=1$, and if $r_1=1$, then $|\tan\alpha|=2$.

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radius of $S_2$ is not required?

The fact that $S_2$ is drawn taking $SQ$ as diameter is required to prove that $|\tan\alpha|=\tan\theta$ as we've already seen.

To find $\tan\theta$, we don't need to use $r_2$ because we can use the fact that $\triangle{PQR}$ is a right triangle and get $\tan\theta=\frac{4}{2r_1}=\frac{2}{r_1}$ to conclude $|\tan\alpha|=\frac{2}{r_1}$.

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In the following, I'm going to add some explanations.

$\triangle{PQR}$ is a right triangle with $PQ=2r_1,RP=4$ and $\angle{PQR}=\theta$ satisfying $\tan\theta=\frac{4}{2r_1}=\frac{2}{r_1}$, i.e. $\theta=\arctan(\frac{2}{r_1})$.

$\triangle{SQP}$ is a right triangle with $SQ=2r_2$ and $\angle{PQS}=\theta$ satisfying $r_2=r_1\cos\theta=\frac{r_1}{\sqrt{1+\tan^2\theta}}=\frac{r_1}{\sqrt{1+(2/r_1)^2}}=\frac{r_1^2}{\sqrt{r_1^2+4}}$.

So, in conclusion, we can say that

• if $r_1=1$, then $|\tan\alpha|=\tan\theta=2,\theta=\arctan(2)$ and $r_2=\frac{1}{\sqrt 5}$.

• if $r_1=2$, then $|\tan\alpha|=\tan\theta=1,\theta=\frac{\pi}{4}$ and $r_2=\sqrt 2$.

Answer 2

I do not see any problem.

Since $(PS)$ is tangent to the circle $\mathcal{S}_2$, the angle between the two circles is also the angle between $\mathcal{S}_2$ and $(PS)$ at $S$ or at $P$, by symmetry. Hence it is the angle between $(PR)$ and $(PS)$. By orthogonality, it is the angle between $(QP)$ and $(QR)$. The tangent of this angle is $PR/PQ = 4/2 = 2$.

Answer 3

Actually, you can prove that $|\tan\theta|=|\tan\alpha|$. The angle between two circles $C_1$ and $C_2$ intersecting at $X$, is defined as the angle between the tangents of $C_1$ and $C_2$ at $X$. Let us denote this angle by $\mu$. Also, let us define the angle between the tangents of $C_1$ and $C_2$ at $X$ by $\nu$. We know that at every point $P$ on the perimeter of a circle $C$, the tangent line and the radius meeting at $P$ are perpendicular. This means that $$ |\mu-\nu|=\text{$0$ or $\pi$} \implies |\tan \mu|=|\tan\nu|. $$ Same situation happens here. By defining the angle between circles $C_1$ and $C_2$ at $S$ as $\alpha$ and the angle between the radii of $C_1$ and $C_2$ at $S$ ($C_1S$ and $QS$) as $\beta$, we see $|\tan\alpha|=|\tan\beta|$. Since $\triangle C_1SQ$ is equilateral, we have $\beta=\measuredangle C_1SQ=\measuredangle C_1QS=\theta$. Therefore, $$ |\tan\alpha|=|\tan \theta|=\frac{PR}{PQ}=2. $$