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(B) Let the centre of the circle $C$ be $(h, k)$. Since the circle passes through $A(2, -1)$ and $B(3, 4)$,the centre $(h, k)$ must lie on the perpend

Vedclass · Accepted Answer

$\frac{65}{2}$

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(B) Let the centre of the circle $C$ be $(h, k)$. Since the circle passes through $A(2, -1)$ and $B(3, 4)$,the centre $(h, k)$ must lie on the perpendicular bisector of $AB$.The midpoint of $AB$ is $M = (\frac{2+3}{2}, \frac{-1+4}{2}) = (\frac{5}{2}, \frac{3}{2})$.The slope of $AB$ is $m_{AB} = \frac{4 - (-1)}{3 - 2} = 5$.The slope of the perpendicular bisector is $m_{\perp} = -\frac{1}{5}$.The equation of the perpendicular bisector is $y - \frac{3}{2} = -\frac{1}{5}(x - \frac{5}{2})$,which simplifies to $x + 5y = 10$.Given that the centre $(h, k)$ lies on the circle $(x-5)^2 + (y-1)^2 = \frac{13}{2}$,we have $(h-5)^2 + (k-1)^2 = \frac{13}{2}$.Also,$h + 5k = 10$,so $h = 10 - 5k$.Substituting $h$ into the circle equation: $(10 - 5k - 5)^2 + (k - 1)^2 = \frac{13}{2} \implies (5 - 5k)^2 + (k - 1)^2 = \frac{13}{2}$.$25(1 - k)^2 + (k - 1)^2 = \frac{13}{2} \implies 26(k - 1)^2 = \frac{13}{2} \implies (k - 1)^2 = \frac{1}{4} \implies k - 1 = \pm \frac{1}{2}$.For $k = \frac{3}{2}$,$h = 10 - 5(\frac{3}{2}) = \frac{5}{2}$. The centre is $(\frac{5}{2}, \frac{3}{2})$,which is the midpoint $M$. This makes $AB$ a diameter,which is excluded.For $k = \frac{1}{2}$,$h = 10 - 5(\frac{1}{2}) = \frac{15}{2}$. The centre is $C = (\frac{15}{2}, \frac{1}{2})$.The radius squared is $r^2 = CA^2 = (\frac{15}{2} - 2)^2 + (\frac{1}{2} - (-1))^2 = (\frac{11}{2})^2 + (\frac{3}{2})^2 = \frac{121}{4} + \frac{9}{4} = \frac{130}{4} = \frac{65}{2}$.

Let $C$ be a circle passing through the points $A (2,-1)$ and $B (3,4)$. The line segment $AB$ is not a diameter of $C$. If $r$ is the radius of $C$ and its centre lies on the circle $(x-5)^{2}+(y-1)^{2}=\frac{13}{2}$,then $r^{2}$ is equal to

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