In $\Delta ABC,A=(1,2);B=(5,5),\angle ACB=90°$. If area of $\Delta ABC$ is to be 6.5 sq. units, the possible number of points for C is

Given that $\left[ABC\right]=\frac{13}{2}$

$\frac{1}{2}\left|\begin{array}{ccc}1& 2& 1\\ 5& 5& 1\\ a& b& 1\end{array}\right|=\frac{13}{2}$

$|(5-b)-2(5-a)+1(5b-5a)|=13$

$-5+4b-3a=13,-13$

$3a-4b+18=0\cdots \cdots \cdots \left(1\right);3a-4b-8=0\cdots \cdots \cdots \left(2\right)$

$A{B}^2=A{C}^2+B{C}^2$

$25={(1-a)}^2+{(2-b)}^2+{(a-5)}^2+{(b-5)}^2$

$2a^2+2b^2-12a-14b+30=0$

$a^2+b^2-6a-7b+15=0$

$CM=\left|\frac{3\left(3\right)-4\left(\frac{7}{2}\right)-8}{5}\right|$

$=\frac{13}{5}=2.6$

$CM>r$

$\&C{M}^1>r$

This is no point for C