If A(-1,3) and B(5,3) are points on a circle | vedclass

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(B) Given $\angle APB = \pi / 4$. The angle subtended by the chord $AB$ at the center $O(h, k)$ is $\angle AOB = 2 \angle APB = \pi / 2$.Since $OA = O

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$x^2+y^2-4x-12y+22=0$

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(B) Given $\angle APB = \pi / 4$. The angle subtended by the chord $AB$ at the center $O(h, k)$ is $\angle AOB = 2 \angle APB = \pi / 2$.Since $OA = OB$,$	riangle OAB$ is an isosceles right-angled triangle.The midpoint of $AB$ is $(\frac{-1+5}{2}, \frac{3+3}{2}) = (2, 3)$. The perpendicular bisector of $AB$ is $x = 2$,so $h = 2$.Since $\angle AOB = 90^\circ$,the distance from $O(2, k)$ to $A(-1, 3)$ is $R$,and $OA^2 + OB^2 = AB^2$.$AB^2 = (5 - (-1))^2 + (3 - 3)^2 = 6^2 = 36$.$OA^2 = (2 - (-1))^2 + (k - 3)^2 = 9 + (k - 3)^2$.Since $OA = OB$,$OA^2 = OB^2 = R^2$,so $2R^2 = 36 \Rightarrow R^2 = 18$.$9 + (k - 3)^2 = 18$ $\Rightarrow (k - 3)^2 = 9$ $\Rightarrow k - 3 = \pm 3$.Thus,$k = 6$ or $k = 0$.The centers are $(2, 6)$ and $(2, 0)$.For center $(2, 6)$,the equation is $(x - 2)^2 + (y - 6)^2 = 18$ $\Rightarrow x^2 - 4x + 4 + y^2 - 12y + 36 = 18$ $\Rightarrow x^2 + y^2 - 4x - 12y + 22 = 0$.For center $(2, 0)$,the equation is $(x - 2)^2 + (y - 0)^2 = 18$ $\Rightarrow x^2 - 4x + 4 + y^2 = 18$ $\Rightarrow x^2 + y^2 - 4x - 14 = 0$.

If $A(-1,3)$ and $B(5,3)$ are points on a circle $C$ and the chord $AB$ subtends an angle $\pi / 4$ at a point $P$ on $C$,then the equation of such a circle $C$ is

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