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# The area of the circle centred at $(1,2)$ and passing through $(4,6)$ is

$\begin{array}{1 1}5\pi\\10\pi\\25\pi\\30\pi\end{array}$

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• General equation of a circle with centre $(h,k)$ is $(x-h)^2+(y-k)^2=a^2$
Answer : $25\pi$
The centre of the circle is $(1,2)$
$(x-1)^2+(y-2)^2=a^2$
It passes through $(4,6)$
$\therefore (4-1)^2+(6-2)^2=a^2$
$\Rightarrow 9+16=a^2$
$\therefore a=5$
$\therefore$ Radius $a=5$
Hence the area of the circle is $\pi a^2=\pi (5)^2$
$\Rightarrow 25\pi$