# Touch x-axis

Find the equations of circles that pass through points A (-2; 4) and B (0; 2) and touch the x-axis.

Result

a = (Correct answer is: a=pow(x+2, 2)+pow(y-2, 2)=4)
b = (Correct answer is: b=pow(x-6, 2)+pow(y-10, 2)=100)

#### Solution:

$(x-m)^2+(y-n)^2 = r^2 \ \\ \ \\ (m+2)^2+(n-4)^2 = r^2 \ \\ m^2+(n-2)^2 = r^2 \ \\ n = r \ \\ \ \\ (m+2)^2+(n-4)^2 = n^2 \ \\ m^2+(n-2)^2 = n^2 \ \\ \ \\ m^2 + 4 \ m - 8 \ n + 20 = 0 \ \\ m^2 - 4 \ n + 4 = 0 \ \\ \ \\ n = (m^2+4)/4 \ \\ \ \\ \ \\ \ \\ m^2 + 4 \ m - 8 \cdot \ ((m^2+4)/4) + 20 = 0 \ \\ -m^2 +4m +12 = 0 \ \\ m^2 -4m -12 = 0 \ \\ \ \\ a = 1; b = -4; c = -12 \ \\ D = b^2 - 4ac = 4^2 - 4\cdot 1 \cdot (-12) = 64 \ \\ D>0 \ \\ \ \\ m_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 4 \pm \sqrt{ 64 } }{ 2 } \ \\ m_{1,2} = \dfrac{ 4 \pm 8 }{ 2 } \ \\ m_{1,2} = 2 \pm 4 \ \\ m_{1} = 6 \ \\ m_{2} = -2 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (m -6) (m +2) = 0 \ \\ n_{ 1 } = (m_{ 1 }^2+4)/4 = (6^2+4)/4 = 10 \ \\ \ \\ n_{ 2 } = (m_{ 2 }^2+4)/4 = ((-2)^2+4)/4 = 2 \ \\ \ \\ r = n \ \\ \ \\ r = 2, m = -2, n = 2 \ \\ r = 10, m = 6, n = 10 \ \\ \ \\ \ \\ a = (x+2)^2+(y-2)^2 = 4$
$b = (x-6)^2+(y-10)^2 = 100$

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#### Following knowledge from mathematics are needed to solve this word math problem:

For Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc. Looking for help with calculating roots of a quadratic equation? Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? Pythagorean theorem is the base for the right triangle calculator.

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