Hi,

I need to solve two quadratics simultaneously to find the x and y values,
they are:

x^2 + y^2 = 0.23
(x-0.15)^2 + (y-0.175)^2 = 0.16

Thank you

$(x - 0.15)^2 + (y - 0.175)^2 = 0.16$

$x^2 - 0.3x + 0.15^2 + y - 0.35y + 0.175^2 = 0.16$

$x^2 + y^2 - 0.3x - 0.35y = 0.16 - 0.15^2 - 0.175^2$

Now substitute in for $x^2 + y^2$:

$0.23 - 0.3x - 0.35y = 0.106875$

$0.3x + 0.35y = 0.123125$

Now solve for y in terms of x and substitute y = f(x) back into the first equation:

$x^2 + (f(x))^2 = 0.23$

Now you have a quadratic equation in terms of x which can be solved using the quadratic formula.

3. Thanks for helping me but when I substitute in the results I get from the quadratic formula into the two equations I do not get the correct result,

Can someone show me how to get to the quadratic formula so I can check my answers please?

4. That should work. How about if YOU show your work and why you think the answer you got was wrong?

5. Forming an equation in y:
0.35y = 0.123125 - 0.3x
y =(0.123125 - 0.3x)/0.35

Substitute into x^2 + y^2 = 0.23 gives:
x^2 + ((0.123125 - 0.3x)/0.35)^2 = 0.23

Expanding brackets:
x^2 + (0.09x^2 - 0.6x + 0.015159765)/(0.35^2) = 0.23

Multiplying both sides by (0.35^2):
0.1225x^2 + 0.09x^2 - 0.6x + 0.015159765 = 0.028175

0.2125x^2 - 0.6x - 0.013015235 = 0

a = 0.2125
b = -0.6
c = -0.013015235

b^2 = 0.36
4ac = -0.011062949
b^2 - 4ac = 0.371062949
sqrt(b^2 - 4ac) = 0.609149365

-b - ans = -0.009149365
-b + ans = 1.209149366

x = 2.845057332
or
x = -0.021527917

Substituting back into first equation:
when x = 2.845057332, y = NaN
when x = -0.021527917, y = 0.4790997272900794

Checking in equation 2:
(-0.021527917 - 0.15)^2 + (0.4790997272900794 - 0.175)^2 = 0.12189847124672402

The answer doesnt equal 0.16, can someone please explain what I have done wrong?