Let the circumcentre of a triangle with vertices A a , 3 , B b , 5 and C a , b , a b > 0 be P 1 , 1 . If…

On plotting the graph we get,

 

Now by diagram $m_{AC}⟶\infty$ and $m_{PD}=0$ as $PD$ is perpendicular bisector, so $D\left(\frac{a+a}{2},\frac{b+3}{2}\right)$

$\equiv D\left(a,\frac{b+3}{2}\right)$

Now since $m_{PD}=0$, so $\frac{b+3}{2}-1=0$

$\Rightarrow b+3-2=0$$\Rightarrow b=-1$

Now similarly finding the coordinates of $E\left(\frac{b+a}{2},\frac{5+b}{2}\right)=\left(\frac{a-1}{2},2\right) \left\{\text{as} b=-1\right\}$

Now we know that product of perpendicular slopes is $m_{CB}·m_{EP}=-1$

$\Rightarrow \left(\frac{5-b}{b-a}\right)=\left(\frac{2-1}{\frac{a-1}{2}-1}\right)=-1$

$\Rightarrow \left(\frac{6}{-1-a}\right)=\left(\frac{2}{a-3}\right)=-1$

$\Rightarrow 12=\left(1+a\right)\left(a-3\right)$

$\Rightarrow 12=a^2-3a+a-3$

$\Rightarrow a^2-2a-15=0$

$\Rightarrow \left(a-5\right)\left(a+3\right)=0$

So, $a=5$ or $a=-3$

Now given $ab>0$

So, $a\left(-1\right)>0$$\Rightarrow a<0$

So, $a=-3$ 

Now equation of line $AP$ by two point form will be,  {where $A\left(-3,3\right) \& P\left(1,1\right)$}

$y-1=\left(\frac{3-1}{-3-1}\right)\left(x-1\right)$

$\Rightarrow -2y+2=x-1$

$\Rightarrow x+2y=3$   $\cdots \left(1\right)$

Now equation of line $BC$ {where $B\left(-1,5\right)$ and $C\left(-3,-1\right)$} will be, 

$\left(y-5\right)=\frac{6}{2}\left(x+1\right)$

$\Rightarrow y-5=3x+3$

$\Rightarrow y=3x+8 \cdots \left(2\right)$

Now solving $\left(1\right)$ and $\left(2\right)$ we get,

$x+2\left(3x+8\right)=3$

$\Rightarrow 7x+16=3$$\Rightarrow 7x=-13$

$\Rightarrow x=-\frac{13}{7}$

And $y=3\left(-\frac{13}{7}\right)+8$$=\frac{-39+56}{7}$

So, $y=\frac{17}{7}$

So the value of $\left(k_1,k_2\right)$ will be $x+y=\frac{-13+17}{7}=\frac{4}{7}$