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If the parametric equations of the circle passing through the points $(3,4),(3,2)$ and $(1,4)$ is $x=a+r \cos \theta, y=b+r \sin \theta$, then $b^a r^a=$
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Verified Answer
The correct answer is:
18
A circle is passing through $(3,4),(3,2)$ and $(1,4)$.
Clearly, $\triangle A B C$ is a right angle triangle
$-$ centre of circle is mid-point of $A C$ i.e., $O(2,3)$
$r=A O=\sqrt{(3-2)^2+(2-3)^2}=\sqrt{1+1}=\sqrt{2}$
Equation of circle in parameter form is
$\begin{aligned}
& x=2+\sqrt{2} \cos \theta \Rightarrow y=3+\sqrt{2} \sin \theta \\
& \therefore \quad a=2, b=3, r=\sqrt{2} \\
& b^a \cdot r^a=(3)^2(\sqrt{2})^2=9 \times 2=18 \\
&
\end{aligned}$
Clearly, $\triangle A B C$ is a right angle triangle
$-$ centre of circle is mid-point of $A C$ i.e., $O(2,3)$
$r=A O=\sqrt{(3-2)^2+(2-3)^2}=\sqrt{1+1}=\sqrt{2}$
Equation of circle in parameter form is
$\begin{aligned}
& x=2+\sqrt{2} \cos \theta \Rightarrow y=3+\sqrt{2} \sin \theta \\
& \therefore \quad a=2, b=3, r=\sqrt{2} \\
& b^a \cdot r^a=(3)^2(\sqrt{2})^2=9 \times 2=18 \\
&
\end{aligned}$
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