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Sorry for a non-answer. Pictures are so beautiful I could not resist and made another illustration. Shown in red is the image under $\lambda$ of the geodesic from $\mathbb H$ that is the semicircle with endpoints $22/17$ and $23/17$.

Here is (part of) the calculation of the $\Phi_{10}$ case from the answer by David E Speyer. As there, the resulting polynomial factors into $9$ irreducibles, $3$ with real coefficients and $3$ complex conjugate pairs. One of the three reals:

$$\scriptscriptstyle{16777216-201326592 x+100663296 \left(11 x^2+y^2\right)-335544320 x \left(11 x^2+3 y^2\right)-31457280 \left(2757 x^4+5898 x^2 y^2+3013 y^4\right) +12582912 x \left(74469 x^4+150090 x^2 y^2+75365 y^4\right)-2097152 \left(2817703 x^6+5080765 x^4 y^2+1703045 x^2 y^4-559505 y^6\right) +16777216 x \left(x^2+y^2\right) \left(1563093 x^4+1102060 x^2 y^2-461225 y^4\right)-327680 \left(x^2+y^2\right)^2 \left(99837117 x^4-381533062 x^2 y^2-159337027 y^4\right)-2621440 x \left(x^2+y^2\right)^2 \left(80168947 x^4+324318310 x^2 y^2+159693043 y^4\right)+262144 \left(x^2+y^2\right)^2 \left(3319391289 x^6+10893265297 x^4 y^2+7934697207 x^2 y^4+1548990879 y^6\right)-524288 x \left(x^2+y^2\right)^2 \left(1803952329 x^6+9504803807 x^4 y^2+11439158627 x^2 y^4+4277013069 y^6\right)-16384 \left(x^2+y^2\right)^2 \left(80616153629 x^8-116483137740 x^6 y^2-471724634834 x^4 y^4-332885326540 x^2 y^6-49133177571 y^8\right)+32768 x \left(x^2+y^2\right)^2 \left(162384970239 x^8+326354545596 x^6 y^2+112042309434 x^4 y^4-118726692036 x^2 y^6-65725684289 y^8\right)-81920 \left(x^2+y^2\right)^3 \left(92443465791 x^8+214662714384 x^6 y^2+140433238710 x^4 y^4+11787598392 x^2 y^6-5352649901 y^8\right)+327680 x \left(x^2+y^2\right)^4 \left(18924971821 x^6+46132456837 x^4 y^2+30488552835 x^2 y^4+3000049451 y^6\right)-1280 \left(x^2+y^2\right)^5 \left(2468136499389 x^6+6067921659703 x^4 y^2+3772079449143 x^2 y^4+131492099517 y^6\right)+1024 x \left(x^2+y^2\right)^6 \left(964753103313 x^4+2235768664994 x^2 y^2+1287936301521 y^4\right)-512 \left(x^2+y^2\right)^7 \left(336466085491 x^4+635946737924 x^2 y^2+306085239441 y^4\right)+5120 x \left(x^2+y^2\right)^8 \left(2374492413 x^2+2302317821 y^2\right)+160 \left(x^2+y^2\right)^9 \left(1129381041 x^2+1254259889 y^2\right)+405291200 x \left(x^2+y^2\right)^{10}-1450080 \left(x^2+y^2\right)^{11}+\left(x^2+y^2\right)^{12} }$$

The curves look like

Consecutively zooming in near zero for the first one:

Sorry for a non-answer. Pictures are so beautiful I could not resist and made another illustration. Shown in red is the image under $\lambda$ of the geodesic from $\mathbb H$ that is the semicircle with endpoints $22/17$ and $23/17$.

Here is (part of) the calculation of the $\Phi_{10}$ case from the answer by David E Speyer. As there, the resulting polynomial factors into $9$ irreducibles, $3$ with real coefficients and $3$ complex conjugate pairs. One of the three reals:

\begin{align*} &\scriptscriptstyle{16777216-201326592 x}\\ &\scriptscriptstyle{+100663296 \left(11 x^2+y^2\right)-335544320 x \left(11 x^2+3 y^2\right)}\\ &\scriptscriptstyle{-31457280 \left(2757 x^4+5898 x^2 y^2+3013 y^4\right)}\\ &\scriptscriptstyle{+12582912 x \left(74469 x^4+150090 x^2 y^2+75365 y^4\right)}\\ &\scriptscriptstyle{-2097152 \left(2817703 x^6+5080765 x^4 y^2+1703045 x^2 y^4-559505 y^6\right)}\\ &\scriptscriptstyle{+16777216 x \left(x^2+y^2\right) \left(1563093 x^4+1102060 x^2 y^2-461225 y^4\right)}\\ &\scriptscriptstyle{-327680 \left(x^2+y^2\right)^2 \left(99837117 x^4-381533062 x^2 y^2-159337027 y^4\right)}\\ &\scriptscriptstyle{-2621440 x \left(x^2+y^2\right)^2 \left(80168947 x^4+324318310 x^2 y^2 +159693043 y^4\right)}\\ &\scriptscriptstyle{+262144 \left(x^2+y^2\right)^2 \left(3319391289 x^6+10893265297 x^4 y^2+7934697207 x^2 y^4+1548990879 y^6\right)}\\ &\scriptscriptstyle{-524288 x \left(x^2+y^2\right)^2 \left(1803952329 x^6+9504803807 x^4 y^2+11439158627 x^2 y^4+4277013069 y^6\right)}\\ &\scriptscriptstyle{-16384 \left(x^2+y^2\right)^2 \left(80616153629 x^8-116483137740 x^6 y^2-471724634834 x^4 y^4-332885326540 x^2 y^6-49133177571 y^8\right)}\\ &\scriptscriptstyle{+32768 x \left(x^2+y^2\right)^2 \left(162384970239 x^8+326354545596 x^6 y^2+112042309434 x^4 y^4-118726692036 x^2 y^6-65725684289 y^8\right)}\\ &\scriptscriptstyle{-81920 \left(x^2+y^2\right)^3 \left(92443465791 x^8+214662714384 x^6 y^2+140433238710 x^4 y^4+11787598392 x^2 y^6-5352649901 y^8\right)}\\ &\scriptscriptstyle{+327680 x \left(x^2+y^2\right)^4 \left(18924971821 x^6+46132456837 x^4 y^2+30488552835 x^2 y^4+3000049451 y^6\right)}\\ &\scriptscriptstyle{-1280 \left(x^2+y^2\right)^5 \left(2468136499389 x^6+6067921659703 x^4 y^2+3772079449143 x^2 y^4+131492099517 y^6\right)}\\ &\scriptscriptstyle{+1024 x \left(x^2+y^2\right)^6 \left(964753103313 x^4+2235768664994 x^2 y^2+1287936301521 y^4\right)}\\ &\scriptscriptstyle{-512 \left(x^2+y^2\right)^7 \left(336466085491 x^4+635946737924 x^2 y^2+306085239441 y^4\right)}\\ &\scriptscriptstyle{+5120 x \left(x^2+y^2\right)^8 \left(2374492413 x^2+2302317821 y^2\right)}\\ &\scriptscriptstyle{+160 \left(x^2+y^2\right)^9 \left(1129381041 x^2+1254259889 y^2\right)}\\ &\scriptscriptstyle{+405291200 x \left(x^2+y^2\right)^{10}-1450080 \left(x^2+y^2\right)^{11}+\left(x^2+y^2\right)^{12}} \end{align*}

The curves look like

Consecutively zooming in near zero for the first one:

Sorry for a non-answer. Pictures are so beautiful I could not resist and made another illustration. Shown in red is the image under $\lambda$ of the geodesic from $\mathbb H$ that is the semicircle with endpoints $22/17$ and $23/17$.

Here is (part of) the calculation of the $\Phi_{10}$ case from the answer by David E Speyer. As there, the resulting polynomial factors into $9$ irreducibles, $3$ with real coefficients and $3$ complex conjugate pairs. One of the three reals:

$$\scriptstyle{16777216-201326592 x+100663296 \left(11 x^2+y^2\right)-335544320 x \left(11 x^2+3 y^2\right)-31457280 \left(2757 x^4+5898 x^2 y^2+3013 y^4\right)+12582912 x \left(74469 x^4+150090 x^2 y^2+75365 y^4\right)-2097152 \left(2817703 x^6+5080765 x^4 y^2+1703045 x^2 y^4-559505 y^6\right)+16777216 x \left(x^2+y^2\right) \left(1563093 x^4+1102060 x^2 y^2-461225 y^4\right)-327680 \left(x^2+y^2\right)^2 \left(99837117 x^4-381533062 x^2 y^2-159337027 y^4\right)-2621440 x \left(x^2+y^2\right)^2 \left(80168947 x^4+324318310 x^2 y^2+159693043 y^4\right)+262144 \left(x^2+y^2\right)^2 \left(3319391289 x^6+10893265297 x^4 y^2+7934697207 x^2 y^4+1548990879 y^6\right)-524288 x \left(x^2+y^2\right)^2 \left(1803952329 x^6+9504803807 x^4 y^2+11439158627 x^2 y^4+4277013069 y^6\right)-16384 \left(x^2+y^2\right)^2 \left(80616153629 x^8-116483137740 x^6 y^2-471724634834 x^4 y^4-332885326540 x^2 y^6-49133177571 y^8\right)+32768 x \left(x^2+y^2\right)^2 \left(162384970239 x^8+326354545596 x^6 y^2+112042309434 x^4 y^4-118726692036 x^2 y^6-65725684289 y^8\right)-81920 \left(x^2+y^2\right)^3 \left(92443465791 x^8+214662714384 x^6 y^2+140433238710 x^4 y^4+11787598392 x^2 y^6-5352649901 y^8\right)+327680 x \left(x^2+y^2\right)^4 \left(18924971821 x^6+46132456837 x^4 y^2+30488552835 x^2 y^4+3000049451 y^6\right)-1280 \left(x^2+y^2\right)^5 \left(2468136499389 x^6+6067921659703 x^4 y^2+3772079449143 x^2 y^4+131492099517 y^6\right)+1024 x \left(x^2+y^2\right)^6 \left(964753103313 x^4+2235768664994 x^2 y^2+1287936301521 y^4\right)-512 \left(x^2+y^2\right)^7 \left(336466085491 x^4+635946737924 x^2 y^2+306085239441 y^4\right)+5120 x \left(x^2+y^2\right)^8 \left(2374492413 x^2+2302317821 y^2\right)+160 \left(x^2+y^2\right)^9 \left(1129381041 x^2+1254259889 y^2\right)+405291200 x \left(x^2+y^2\right)^{10}-1450080 \left(x^2+y^2\right)^{11}+\left(x^2+y^2\right)^{12}}$$

The curves look like

Consecutively zooming in near zero for the first one:

Sorry for a non-answer. Pictures are so beautiful I could not resist and made another illustration. Shown in red is the image under $\lambda$ of the geodesic from $\mathbb H$ that is the semicircle with endpoints $22/17$ and $23/17$.

Here is (part of) the calculation of the $\Phi_{10}$ case from the answer by David E Speyer. As there, the resulting polynomial factors into $9$ irreducibles, $3$ with real coefficients and $3$ complex conjugate pairs. One of the three reals:

$$\scriptscriptstyle{16777216-201326592 x+100663296 \left(11 x^2+y^2\right)-335544320 x \left(11 x^2+3 y^2\right)-31457280 \left(2757 x^4+5898 x^2 y^2+3013 y^4\right) +12582912 x \left(74469 x^4+150090 x^2 y^2+75365 y^4\right)-2097152 \left(2817703 x^6+5080765 x^4 y^2+1703045 x^2 y^4-559505 y^6\right) +16777216 x \left(x^2+y^2\right) \left(1563093 x^4+1102060 x^2 y^2-461225 y^4\right)-327680 \left(x^2+y^2\right)^2 \left(99837117 x^4-381533062 x^2 y^2-159337027 y^4\right)-2621440 x \left(x^2+y^2\right)^2 \left(80168947 x^4+324318310 x^2 y^2+159693043 y^4\right)+262144 \left(x^2+y^2\right)^2 \left(3319391289 x^6+10893265297 x^4 y^2+7934697207 x^2 y^4+1548990879 y^6\right)-524288 x \left(x^2+y^2\right)^2 \left(1803952329 x^6+9504803807 x^4 y^2+11439158627 x^2 y^4+4277013069 y^6\right)-16384 \left(x^2+y^2\right)^2 \left(80616153629 x^8-116483137740 x^6 y^2-471724634834 x^4 y^4-332885326540 x^2 y^6-49133177571 y^8\right)+32768 x \left(x^2+y^2\right)^2 \left(162384970239 x^8+326354545596 x^6 y^2+112042309434 x^4 y^4-118726692036 x^2 y^6-65725684289 y^8\right)-81920 \left(x^2+y^2\right)^3 \left(92443465791 x^8+214662714384 x^6 y^2+140433238710 x^4 y^4+11787598392 x^2 y^6-5352649901 y^8\right)+327680 x \left(x^2+y^2\right)^4 \left(18924971821 x^6+46132456837 x^4 y^2+30488552835 x^2 y^4+3000049451 y^6\right)-1280 \left(x^2+y^2\right)^5 \left(2468136499389 x^6+6067921659703 x^4 y^2+3772079449143 x^2 y^4+131492099517 y^6\right)+1024 x \left(x^2+y^2\right)^6 \left(964753103313 x^4+2235768664994 x^2 y^2+1287936301521 y^4\right)-512 \left(x^2+y^2\right)^7 \left(336466085491 x^4+635946737924 x^2 y^2+306085239441 y^4\right)+5120 x \left(x^2+y^2\right)^8 \left(2374492413 x^2+2302317821 y^2\right)+160 \left(x^2+y^2\right)^9 \left(1129381041 x^2+1254259889 y^2\right)+405291200 x \left(x^2+y^2\right)^{10}-1450080 \left(x^2+y^2\right)^{11}+\left(x^2+y^2\right)^{12} }$$

The curves look like

Consecutively zooming in near zero for the first one:

Sorry for a non-answer. Pictures are so beautiful I could not resist and made another illustration. Shown in red is the image under $\lambda$ of the geodesic from $\mathbb H$ that is the semicircle with endpoints $22/17$ and $23/17$.

Conjecture.

The image of a geodesic (either vertical ray or semicircle) is algebraic if and only if the endpoint of the ray, resp. the endpoints of the semicircle are rational.

Later

In view of the counterexample given by David E Speyer and the correction by Ian Agol, the conjecture is not true as stated. Should be rational or conjugate quadratic irrationalities (not sure about the vertical geodesic though).

Sorry for a non-answer. Pictures are so beautiful I could not resist and made another illustration. Shown in red is the image under $\lambda$ of the geodesic from $\mathbb H$ that is the semicircle with endpoints $22/17$ and $23/17$.

Here is (part of) the calculation of the $\Phi_{10}$ case from the answer by David E Speyer. As there, the resulting polynomial factors into $9$ irreducibles, $3$ with real coefficients and $3$ complex conjugate pairs. One of the three reals:

$$\scriptstyle{16777216-201326592 x+100663296 \left(11 x^2+y^2\right)-335544320 x \left(11 x^2+3 y^2\right)-31457280 \left(2757 x^4+5898 x^2 y^2+3013 y^4\right)+12582912 x \left(74469 x^4+150090 x^2 y^2+75365 y^4\right)-2097152 \left(2817703 x^6+5080765 x^4 y^2+1703045 x^2 y^4-559505 y^6\right)+16777216 x \left(x^2+y^2\right) \left(1563093 x^4+1102060 x^2 y^2-461225 y^4\right)-327680 \left(x^2+y^2\right)^2 \left(99837117 x^4-381533062 x^2 y^2-159337027 y^4\right)-2621440 x \left(x^2+y^2\right)^2 \left(80168947 x^4+324318310 x^2 y^2+159693043 y^4\right)+262144 \left(x^2+y^2\right)^2 \left(3319391289 x^6+10893265297 x^4 y^2+7934697207 x^2 y^4+1548990879 y^6\right)-524288 x \left(x^2+y^2\right)^2 \left(1803952329 x^6+9504803807 x^4 y^2+11439158627 x^2 y^4+4277013069 y^6\right)-16384 \left(x^2+y^2\right)^2 \left(80616153629 x^8-116483137740 x^6 y^2-471724634834 x^4 y^4-332885326540 x^2 y^6-49133177571 y^8\right)+32768 x \left(x^2+y^2\right)^2 \left(162384970239 x^8+326354545596 x^6 y^2+112042309434 x^4 y^4-118726692036 x^2 y^6-65725684289 y^8\right)-81920 \left(x^2+y^2\right)^3 \left(92443465791 x^8+214662714384 x^6 y^2+140433238710 x^4 y^4+11787598392 x^2 y^6-5352649901 y^8\right)+327680 x \left(x^2+y^2\right)^4 \left(18924971821 x^6+46132456837 x^4 y^2+30488552835 x^2 y^4+3000049451 y^6\right)-1280 \left(x^2+y^2\right)^5 \left(2468136499389 x^6+6067921659703 x^4 y^2+3772079449143 x^2 y^4+131492099517 y^6\right)+1024 x \left(x^2+y^2\right)^6 \left(964753103313 x^4+2235768664994 x^2 y^2+1287936301521 y^4\right)-512 \left(x^2+y^2\right)^7 \left(336466085491 x^4+635946737924 x^2 y^2+306085239441 y^4\right)+5120 x \left(x^2+y^2\right)^8 \left(2374492413 x^2+2302317821 y^2\right)+160 \left(x^2+y^2\right)^9 \left(1129381041 x^2+1254259889 y^2\right)+405291200 x \left(x^2+y^2\right)^{10}-1450080 \left(x^2+y^2\right)^{11}+\left(x^2+y^2\right)^{12}}$$

The curves look like

Consecutively zooming in near zero for the first one: