# 4 circles and 3 tangents

• Aug 27th 2009, 07:47 PM
dhiab
4 circles and 3 tangents
Find equations for 4 circles each has 3 tangents ,equation of the tangents are :

$\displaystyle \begin{array}{l} y = - \frac{3}{4}x - 4 \\ y = \frac{2}{5}x + 4 \\ y = x + 1 \\ \end{array}$
• Aug 27th 2009, 11:32 PM
red_dog
The three tangents are:

$\displaystyle d_1: 3x+4y+16=0$

$\displaystyle d_2: 2x-5y+20=0$

$\displaystyle d_3: x-y+1=0$

Let $\displaystyle C(a,b)$ be the center of one circle. Then $\displaystyle d(d_1,C)=d(d_2,C)=d(d_3,C)$

$\displaystyle \frac{|3a+4b+16|}{5}=\frac{|2a-5b+20|}{\sqrt{29}}=\frac{|a-b+1|}{\sqrt{2}}$

We get four systems:

$\displaystyle \left\{\begin{array}{ll}(3\sqrt{29}-10)a+(4\sqrt{29}+25)+16\sqrt{29}-100=0\\(3\sqrt{2}-5)a+(4\sqrt{2}+5)b+16\sqrt{2}-5=0\end{array}\right.$

$\displaystyle \left\{\begin{array}{ll}(3\sqrt{29}-10)a+(4\sqrt{29}+25)+16\sqrt{29}-100=0\\(3\sqrt{2}+5)a+(4\sqrt{2}-5)b+16\sqrt{2}+5=0\end{array}\right.$

$\displaystyle \left\{\begin{array}{ll}(3\sqrt{29}+10)a+(4\sqrt{2 9}-25)+16\sqrt{29}+100=0\\(3\sqrt{2}-5)a+(4\sqrt{2}+5)b+16\sqrt{2}-5=0\end{array}\right.$

$\displaystyle \left\{\begin{array}{ll}(3\sqrt{29}+10)a+(4\sqrt{2 9}-25)+16\sqrt{29}+100=0\\(3\sqrt{2}+5)a+(4\sqrt{2}-5)b+16\sqrt{2}+5=0\end{array}\right.$

Solving the systems we get the centers of the four circles. Then find their radius and equation.