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:

$$\begin{gathered} (x-m{)}^2+(y-n{)}^2=r^2 \\ (m+2{)}^2+(n-4{)}^2=r^2 \\ {m}^2+(n-2{)}^2=r^2 \\ n=r \\ (m+2{)}^2+(n-4{)}^2=n^2 \\ {m}^2+(n-2{)}^2=n^2 \\ {m}^2+4m-8n+20=0 \\ {m}^2-4n+4=0 \\ n=(m^2+4)\mathrm{/}4 \\ {m}^2+4m-8\ast ((m^2+4)\mathrm{/}4)+20=0 \\ {m}^2+4m-8\cdot ((m^2+4)\mathrm{/}4)+20=0 \\ -m^2+4m+12=0 \\ {m}^2-4m-12=0 \\ a=1;b=-4;c=-12 \\ D=b^2-4ac=4^2-4\cdot 1\cdot (-12)=64 \\ D>0 \\ {m}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{4\pm \sqrt{64}}{2}} \\ {m}_{1,2}={\displaystyle \frac{4\pm 8}{2}} \\ {m}_{1,2}=2\pm 4 \\ {m}_1=6 \\ {m}_2=-2 \\ \text{ Factored form of the equation: } \\ (m-6)(m+2)=0 \\ {n}_1=(m_1^2+4)\mathrm{/}4=(6^2+4)\mathrm{/}4=10 \\ {n}_2=(m_2^2+4)\mathrm{/}4=((-2{)}^2+4)\mathrm{/}4=2 \\ r=n \\ r=2,m=-2,n=2 \\ r=10,m=6,n=10 \\ a=(x+2{)}^2+(y-2{)}^2=4 \end{gathered}$$

$b=(x-6{)}^2+(y-10{)}^2=100$

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Showing 1 comment:

Math student

how do you end up with n=r ?

8 months ago  Like

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