**Step 1:** Let’s locate the formula of the usual chord by deducting the formulas of the circles:

$ latex x ^ 2+ y ^ 2-4x +3 y +5= 0$

$ latex x ^ 2+ y ^ 2-6x +5 y +9= 0 ~(-$

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$ latex 2x-2y-4= 0$

**Step 2:** We can streamline the formula acquired by splitting by 2, as well as fixing for

*x*, we have:

$ latex x-y-2= 0$

$ latex x= y-2$

**Step 3:** Let’s replacement the formula from action 2 right into the formula of the initial circle. After that, we have:

$ latex x ^ 2+ y ^ 2-4x +3 y +5= 0$

$$( y-2) ^ 2+ y ^ 2 +4( y-2) +3 y +5= 0$$

$ latex 2y ^ 2 +3 y +1= 0$

$ latex (2y +1)( y +1)= 0$

Solving, we have $latex y=- frac {1} {2} $ as well as $latex y= -1$.

** Action 4:** When $latex y=- frac {1} {2} $, we have $latex x= 1 ~ frac {1} {2} $ as well as when $latex y= -1$, we have $latex x= 1$.

As a result, the factors of junction of the circles are $latex (1 ~ frac {1} {2}, ~- frac {1} {2} )$ as well as $latex (1, -1)$.