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(B) Let the points be $A(3,2)$,$B(3,4)$,and $C(1,4)$.Since the line segment $AB$ is vertical $(x=3)$ and $BC$ is horizontal $(y=4)$,the angle $\angle

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) Let the points be $A(3,2)$,$B(3,4)$,and $C(1,4)$.Since the line segment $AB$ is vertical $(x=3)$ and $BC$ is horizontal $(y=4)$,the angle $\angle ABC = 90^\circ$.Thus,$AC$ is the diameter of the circle.The center of the circle $O(a,b)$ is the midpoint of $AC$:$a = \frac{3+1}{2} = 2$,$b = \frac{2+4}{2} = 3$.The radius $r$ is the distance from the center $(2,3)$ to any point,e.g.,$A(3,2)$:$r = \sqrt{(3-2)^2 + (2-3)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.We need to find $b^a r^a$:$b^a r^a = 3^2 \cdot (\sqrt{2})^2 = 9 \cdot 2 = 18$.

If the parametric equations of the circle passing through the points $(3,4), (3,2)$ and $(1,4)$ are $x=a+r \cos \theta$ and $y=b+r \sin \theta$,then find the value of $b^a r^a$.

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