The vertices of a triangle are given as A(2,1), B(2,4) and C(4,1). The triangle undergoes a rotation to give an image with vertices A'(2,1), B'(1,2) and C'(3,2).

Kituo cha mzunguko kiko kwenye bisector ya perpendicular ya kila sehemu inayounganisha pointi na picha yake.

Kwa AA': A(2,1) na A'(5,1)
Midpoint $M_{AA'} = \left(\frac{2+5}{2}, \frac{1+1}{2}\right) = \left(\frac{7}{2}, 1\right) = (3.5, 1)$.
Mteremko wa AA' ni $m_{AA'} = \frac{1-1}{5-2} = 0$. Hii ni mstari mlalo.
Bisector ya perpendicular itakuwa mstari wima unaopita katikati ya $M_{AA'}$, kwa hivyo equation yake ni $x = 3.5$.

Kwa BB': B(2,4) na B'(3,1)
Midpoint $M_{BB'} = \left(\frac{2+3}{2}, \frac{4+1}{2}\right) = \left(\frac{5}{2}, \frac{5}{2}\right) = (2.5, 2.5)$.
Mteremko wa BB' ni $m_{BB'} = \frac{1-4}{3-2} = \frac{-3}{1} = -3$.
Mteremko wa bisector ya perpendicular ni $m_{\perp BB'} = -\frac{1}{-3} = \frac{1}{3}$.
Equation ya bisector ya perpendicular ni $y - 2.5 = \frac{1}{3}(x - 2.5)$.
$y = \frac{1}{3}x - \frac{2.5}{3} + 2.5$
$y = \frac{1}{3}x - \frac{5}{6} + \frac{15}{6}$
$y = \frac{1}{3}x + \frac{10}{6}$
$y = \frac{1}{3}x + \frac{5}{3}$.

Ili kupata kituo cha mzunguko, tunasuluhisha mifumo ya equations:
$x = 3.5$
$y = \frac{1}{3}x + \frac{5}{3}$
Badilisha $x = 3.5 = \frac{7}{2}$ kwenye equation ya pili:
$y = \frac{1}{3}\left(\frac{7}{2}\right) + \frac{5}{3} = \frac{7}{6} + \frac{10}{6} = \frac{17}{6}$.
Kwa hivyo, kituo cha mzunguko kulingana na pointi A na B ni $P\left(3.5, \frac{17}{6}\right)$.

Kituo cha mzunguko ni $P\left(\frac{7}{2}, \frac{17}{6}\right)$.
Vector PA ni $\vec{PA} = \left(2 - \frac{7}{2}, 1 - \frac{17}{6}\right) = \left(\frac{4-7}{2}, \frac{6-17}{6}\right) = \left(-\frac{3}{2}, -\frac{11}{6}\right)$.
Vector PA' ni $\vec{PA'} = \left(5 - \frac{7}{2}, 1 - \frac{17}{6}\right) = \left(\frac{10-7}{2}, \frac{6-17}{6}\right) = \left(\frac{3}{2}, -\frac{11}{6}\right)$.

Urefu wa vectors hizi:
$|\vec{PA}| = \sqrt{\left(-\frac{3}{2}\right)^2 + \left(-\frac{11}{6}\right)^2} = \sqrt{\frac{9}{4} + \frac{121}{36}} = \sqrt{\frac{81+121}{36}} = \sqrt{\frac{202}{36}} = \frac{\sqrt{202}}{6}$.
$|\vec{PA'}| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(-\frac{11}{6}\right)^2} = \sqrt{\frac{9}{4} + \frac{121}{36}} = \sqrt{\frac{81+121}{36}} = \sqrt{\frac{202}{36}} = \frac{\sqrt{202}}{6}$.
Urefu ni sawa, ambayo inathibitisha P ni kituo cha mzunguko kwa A na A'.

Dot product $\vec{PA} \cdot \vec{PA'} = \left(-\frac{3}{2}\right)\left(\frac{3}{2}\right) + \left(-\frac{11}{6}\right)\left(-\frac{11}{6}\right) = -\frac{9}{4} + \frac{121}{36} = \frac{-81+121}{36} = \frac{40}{36} = \frac{10}{9}$.
$\cos\theta = \frac{\vec{PA} \cdot \vec{PA'}}{|\vec{PA}| |\vec{PA'}|} = \frac{\frac{10}{9}}{\left(\frac{\sqrt{202}}{6}\right)^2} = \frac{\frac{10}{9}}{\frac{202}{36}} = \frac{10}{9} \times \frac{36}{202} = \frac{40}{202} = \frac{20}{101}$.
$\theta = \arccos\left(\frac{20}{101}\right) \approx 78.5^\circ$.

Ili kuamua mwelekeo (clockwise au counter-clockwise), tunatumia determinant:
$\det(\vec{PA}, \vec{PA'}) = (-\frac{3}{2})(-\frac{11}{6}) - (-\frac{11}{6})(\frac{3}{2}) = \frac{33}{12} - (-\frac{33}{12}) = \frac{66}{12} = \frac{11}{2} > 0$.
Kwa kuwa determinant ni chanya, mzunguko ni counter-clockwise.

Kituo cha mzunguko ni $P\left(\frac{7}{2}, \frac{17}{6}\right)$. Angle ya mzunguko ni $\theta = \arccos\left(\frac{20}{101}\right)$ counter-clockwise.
Pointi C ni $(4,4)$.
Vector PC ni $\vec{PC} = \left(4 - \frac{7}{2}, 4 - \frac{17}{6}\right) = \left(\frac{8-7}{2}, \frac{24-17}{6}\right) = \left(\frac{1}{2}, \frac{7}{6}\right)$.
Urefu $|\vec{PC}| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{7}{6}\right)^2} = \sqrt{\frac{1}{4} + \frac{49}{36}} = \sqrt{\frac{9+49}{36}} = \sqrt{\frac{58}{36}} = \frac{\sqrt{58}}{6}$.

Sasa tunazungusha $\vec{PC}$ kwa angle $\theta$ counter-clockwise.
Ikiwa $\vec{v} = (x,y)$ inazungushwa hadi $\vec{v'} = (x',y')$ karibu na asili, basi $x' = x\cos\theta - y\sin\theta$ na $y' = x\sin\theta + y\cos\theta$.
Hapa, $\cos\theta = \frac{20}{101}$.
$\sin\theta = \sqrt{1 - \left(\frac{20}{101}\right)^2} = \sqrt{\frac{101^2 - 20^2}{101^2}} = \sqrt{\frac{10201 - 400}{10201}} = \sqrt{\frac{9801}{10201}} = \frac{99}{101}$.

Tuna $\vec{PC} = \left(\frac{1}{2}, \frac{7}{6}\right)$.
$x' = \frac{1}{2}\left(\frac{20}{101}\right) - \frac{7}{6}\left(\frac{99}{101}\right) = \frac{10}{101} - \frac{7 \times 33}{2 \times 101} = \frac{10}{101} - \frac{231}{202} = \frac{20 - 231}{202} = -\frac{211}{202}$.
$y' = \frac{1}{2}\left(\frac{99}{101}\right) + \frac{7}{6}\left(\frac{20}{101}\right) = \frac{99}{202} + \frac{140}{606} = \frac{297 + 140}{606} = \frac{437}{606}$.
Hivyo, $\vec{PC''} = \left(-\frac{211}{202}, \frac{437}{606}\right)$.

Pointi C'' (picha ya C) itakuwa $P + \vec{PC''}$:
$C'' = \left(\frac{7}{2} - \frac{211}{202}, \frac{17}{6} + \frac{437}{606}\right) = \left(\frac{707 - 211}{202}, \frac{1717 + 437}{606}\right) = \left(\frac{496}{202}, \frac{2154}{606}\right) = \left(\frac{248}{101}, \frac{359}{101}\right)$.
$C'' \approx (2.455, 3.554)$.
Pointi C' iliyotolewa ni $(3,-2)$.
Kuratibu hizi hazilingani. Hii inathibitisha kuwa data iliyotolewa haielezei mzunguko safi.

Kwa kuwa swali linasisitiza kuwa ni mzunguko, na kwa kuwa bisector za perpendicular za AA' na CC' zinakutana kwa $(3.5, 1)$, na bisector za perpendicular za AA' na BB' zinakutana kwa $(3.5, \frac{17}{6})$, kuna uwezekano mkubwa wa makosa katika kuratibu za swali.

Ikiwa tungechukua kituo cha mzunguko kuwa $(3.5, 1)$ (kama inavyoonekana kutoka kwa AA' na CC'), basi:
Vector PA = $(2-3.5, 1-1) = (-1.5, 0)$.
Vector PA' = $(5-3.5, 1-1) = (1.5, 0)$.
Hii inaonyesha mzunguko wa $180^\circ$ karibu na $(3.5, 1)$ kwa pointi A.
Hebu tuangalie C:
Vector PC = $(4-3.5, 4-1) = (0.5, 3)$.
Vector PC' = $(3-3.5, -2-1) = (-0.5, -3)$.
Hii pia inaonyesha mzunguko wa $180^\circ$ karibu na $(3.5, 1)$ kwa pointi C.
Sasa hebu tuangalie B:
Vector PB = $(2-3.5, 4-1) = (-1.5, 3)$.
Ikiwa tunazungusha PB kwa $180^\circ$ karibu na $(3.5, 1)$, tunapata $B'' = (3.5 - (-1.5), 1 - 3) = (5, -2)$.
Lakini B' iliyotolewa ni $(3,1)$. Hii hailingani.

Kwa hivyo, hakuna mzunguko mmoja unaolingana na pointi zote tatu.
Hata hivyo, ikiwa tunapaswa kutoa jibu, tunaweza kudhani kuwa kuna makosa katika B' na kwamba mzunguko ni $180^\circ$ karibu na $(3.5, 1)$.

Jibu la mwisho kulingana na dhana ya makosa katika swali:
Kutokana na uchambuzi wa bisector za perpendicular, pointi zilizotolewa hazilingani na mzunguko mmoja safi. Hata hivyo, ikiwa tunachukua pointi A na C, kituo cha mzunguko kinapatikana kuwa $(3.5, 1)$ na angle ya mzunguko ni $180^\circ$. Pointi B haipatani na mzunguko huu.

Kituo cha mzunguko: $\boxed{\left(3.5, 1\right)}$
Angle ya mzunguko: $\boxed{180^\circ}$