How to find the center of a circle given the three points (-3,5); (3,3); (11,19)?

Step 1: Given that:
The given function is
$y=Asi{n}^{2}x$

Given range, $x=0\to x=\pi$
Step 2: Calculation of the average value of the given function:
Average value of a function$(<y>)=\frac{\int ydx}{\int dx}$
Now, for a given range, the average value will be,

$<y>=\frac{{\int }_0^{\pi }A{\sin }^{2}xdx}{{\int }_0^{\pi }dx}$
$<y>=\frac{A{\int }_O^{\pi }\frac{1-\cos 2x}{2}dx}{{\int }_0^{\pi }{x}^{O}dx}$
$<y>=\frac{A({\int }_O^{\pi }\frac{1}{2}dx-\frac{1}{2}{\int }_O^{\pi }\cos 2xdx)}{[x_O^{\pi }]}$
Now, $\int \cos nxdx=\frac{1}{n}\sin nx$
$<y>=\frac{A(\frac{1}{2}(x_O^{\pi })-\frac{1}{2}(\frac{1}{2}\cos 2x_O^{\pi }))}{[x_O^{\pi }]}$
$<y>=A[\frac{\frac{1}{2}(\pi -O)-\frac{1}{4}(\cos 2\pi -\cos O)}{\pi -O}]$
Since, $\cos 2\pi =\cos O=1<y>=A[\frac{\frac{\pi }{2}}{\pi }]$
$<y>=\frac{A}{2}$

Hence, the given function's average value in the given range will be
$\frac{A}{2}$
.

Equal distances separate the points from the center (a, b). Using the distance squares as an equation,
${\displaystyle {(a+3)}^2+{(b-5)}^2={(a-3)}^2+{(b-3)}^2={(a-11)}^2+{(b-19)}^2}$,
From the first two,
${\displaystyle 3a-b=4}$, and from first and the third,
${\displaystyle a+b=16}$.
solving, (a, b) = (3, 13)...