# Quadratic equation need to solve

There are these three quadratic equations:

$(x - x_1)^2 + (y - y_1)^2 + P_1 \cdot h^2 = 0$ ……..(1)

$(x - x_2)^2 + (y - y_2)^2 + P_2 \cdot h^2 = 0$ ……..(2)

$(x - x_3)^2 + (y - y_3)^2 + P_3 \cdot h^2 = 0$ ……..(3)

where we know the values of $x_1, x_2, x_3, y_1, y_2, y_3, P_1, P_2, P_3$.

How can I find $(x,y,h)$?

• (x-x1)2$=(x-x_1)^2$? Also, P1*h2=$P_1h^2$? Jan 8, 2014 at 4:51
• The question's format makes it pretty difficult to understand what's going on, but it seems to be three circles...and then the question makes no much sense, unless the intention here is to find intersection points...or something. Jan 8, 2014 at 4:51
• Antonio, you are right. sorry for format can you help me to solve Jan 8, 2014 at 4:53
• Somewhere in the site there are directions how to properly write mathematics here, but I've no idea where. Better, get into other questions and make "edit" to see the symbols used there. Jan 8, 2014 at 4:55
• You haven't yet said what you want, @user119767...not to mention you haven't shown any self work. Do you want to find intersection points between any two of the given circles or what ? Jan 8, 2014 at 4:56

Subtract (1) from (2) to get a linear equation in $x,y$

Subtract (1) from (3) to get another

From these solve for $x$ and $y$.

Substitute back in (1) to get a quadratic in $h^2$.

Solve to get $h^2$

Note till the very end, you can call $z=h^2$ and finally solve for $h$

Here is a worked out example (the following is computer generated so excuse the poor formatting).

Let the equations be $$\left(y-6\right)^2+\left(x-9\right)^2-{{5\,h^2}\over{64}} \tag 1$$ $$\left(y-9\right)^2+\left(x-19\right)^2-{{61\,h^2}\over{32}} \tag2$$ $$\left( y-7\right)^2+\left(x-1\right)^2-{{25\,h^2}\over{32}} \tag 3$$

Subtract (1) from (2) and also (1) from (3) and simplify to get $$-384\,y-1280\,x-117\,h^2+20800 =0 \\ -128\,y+1024\,x-45\,h^2- 4288 =0$$

Solving the two equations, letting $z=h^2$ we get $$x={{9\,z+16832}\over{2176}} , ~~~y=-{{693\,z-61760}\over{2176}} \tag 4$$ Now substitute in (1) to get $$240165\,z^2-33961600\,z+1189826560=0$$ Solving $$z={{3718208}\over{48033}} , z=64$$ which gives $$h = \pm {{8\,\sqrt{58097}}\over{9\,\sqrt{593}}}, h=\pm 8$$ Substituting in (4) we get $$x={{42992}\over{5337}} , y={{19904}\over{5337}} , h=- {{8\,\sqrt{58097}}\over{9\,\sqrt{593}}} \\ x={{42992 }\over{5337}} , y={{19904}\over{5337}} h={{8\,\sqrt{58097}}\over{9 \,\sqrt{593}}} \\x=8 , y=8 , h=8 \\ x= 8 , y=8 , h=-8$$

• i could not able to find x and y because the two linear equations look like [ A+ xB+ yC + Dh^2= 0 and A1+ xB1+ y*C1+D1*h^2=0 ] where A,B,C,D,A1,B1,C1,D1 are constant then how can i find x and y , could you please help Jan 8, 2014 at 5:40
• @user119767. No, just do what user119767 suggested considering "h" as a constant. Doing what is suggested, you have two linear equations in "x" and "y"; from there "x" and "y" are expressed as a function of all parameters and "h". What has been suggested to you is an elimination procedure which allows you to reduce the problem to one single equation in "h". Jan 8, 2014 at 6:38
• @ClaudeLeibovici. i mentioned in my comments what types of equation i find ''[ A+ xB+ yC + Dh^2= 0 and A1+ xB1+ y*C1+D1*h^2=0 ]'' from this equation how can we find out x and y,as a function of all parameters(we can't do --express x in term of y and h /// express y in terms of x and h) and "h". Jan 8, 2014 at 6:54
• @user119767. Write your equations as B * x + C * y = K. Solve for x and y. Since K is a function of h, inject the expressions of x and y as functions of K in the first equation which only involves now h and you are done. Jan 8, 2014 at 7:06
• any one can tell me which numerical methods, i can use to solve these equation like least square estimation, newton raphson method....etc because i need to use these equation for my simulation, need to find out error Jan 14, 2014 at 8:51