How to find the intersection point of two circle ?
Pleas explain with an example.
Thank you

Dear student, 

$$\begin{gathered} \mathrm{Let} \mathrm{we} \mathrm{take} \mathrm{two} \mathrm{equations} \mathrm{of} \mathrm{circles}: \\ {\mathrm{x}}^2-4\mathrm{x}+4+{\mathrm{y}}^2-6\mathrm{y}+9=16 .....\left(1\right) \\ \mathrm{and} {\mathrm{x}}^2-2\mathrm{x}+{\mathrm{y}}^2+2\mathrm{y}+1=9 .....\left(2\right) \\ \mathrm{Multiplying} \mathrm{all} \mathrm{the} \mathrm{terms} \mathrm{in} \mathrm{the} 1 \mathrm{by} -1 \mathrm{we} \mathrm{get} \\ -{\mathrm{x}}^2+4\mathrm{x}-4-{\mathrm{y}}^2+6\mathrm{y}-9=-16 \\ \mathrm{and} {\mathrm{x}}^2-2\mathrm{x}+{\mathrm{y}}^2+2\mathrm{y}+1=9 \\ \mathrm{Adding} \mathrm{the} \mathrm{above} \mathrm{two} \mathrm{equations} \mathrm{we} \mathrm{get} \\ 2\mathrm{x}+8\mathrm{y}-12=-7 \\ 2\mathrm{x}+8\mathrm{y}=5 \\ \mathrm{x}=\frac{5-8\mathrm{y}}{2} \\ \mathrm{Substituting} \mathrm{in} \left(1\right) \mathrm{we} \mathrm{get}, \\ {\left(\frac{5-8\mathrm{y}}{2}\right)}^2-4\left(\frac{5-8\mathrm{y}}{2}\right) +4+{\mathrm{y}}^2-6\mathrm{y}+9=16 \end{gathered}$$
Quadratic equation in y will come out.
Solve for y and substitute in value of x to get the value of x. 
Value of x and y is the intersection point of the two circles.

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