# 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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Math student
how do you end up with n=r ?

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