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Where the equation of two circle: ${X_a}^2 + (y_a - p)^2 = {r_a}^2$

${X_b}^2 + (y_b - p)^2 = {r_b}^2$

I want to know how to get $(X_c, Y_c)$.

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    Well, notice that $(X_c,Y_c)$ has to satisfy the equation of the first, as well as the second, circle, because $(X_c,Y_c)$ belongs to both (it is their intersection). Unless you have any aditional conditions, you should find two possible solutions, one for each intersection point as shown in the diagram. – 2017-01-24
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    Your equations for the circles are not correct. For the first it is $(x-X_a)^2+(y-Y_a)^2=r_a^2$ Also note that you seem to assume $Y_a$ and $y_a$ are the same. Usually they are different but related if one is capital and one lower case. – 2017-01-24

3 Answers 3

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If you take the two equations, set them both equal to zero. $$X_a^2+(y_a−p)^2-r_a^2=0$$ $$X_b^2+(y_b−p)^2-r_b^2=0$$ Then subtract them from one another you will get a linear equation which passes through both intersection points. You can then set that equal to one of the circle equations to get a quadratic for the intersections.

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This is still a system of equations. $$\begin{cases} (x-h_1)^2+(y-k_1)^2=r_1^2\\ (x-h_2)^2+(y-k_2)^2=r_2^2 \end{cases}$$ If you expand it out and subtract them from each other, you can get a linear equation. You can then solve that linear equation for one variable and substitute it into either of the equations, and you end up with a quadratic that you can solve with the quadratic method.

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Hint: Let $(0,0)$ be the center of the first circle. So we need to solve system of two equations: $$ \left\{ \begin{array}{c} x^2+y^2=r_1^2 \\ (x-x_2)^2+(y-y_2)^2 = r_2^2 \\ \end{array} \right. $$ Subtract the second equation from the first: $$ \left\{ \begin{array}{c} x^2+y^2=r_1^2 \\ -2x_2x - 2y_2y + (x_2^2+y_2^2+r_1^2-r_2^2) = 0\\ \end{array} \right. $$ Thus, we need to calculate intersection of the first circle and the line $Ax+By+C=0$ where $A = -2x_2, B = -2y_2 , C = x_2^2+y_2^2+r_1^2-r_2^2$