# 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:

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