# equation of circle

• Aug 19th 2009, 07:44 AM
khotso
equation of circle
I was given this problem last week as a homework.I tried to work it but with avail:
find the eguation of the circle tangent to the two lines x-3y=-9 and
3x+y-3=0.thanks for the help I will get
• Aug 19th 2009, 10:03 AM
red_dog
$d_1:x-3y+9=0, \ d_2:3x+y-3=0$

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

$\frac{|a-3b+9|}{\sqrt{10}}=\frac{|3a+b-3|}{\sqrt{10}}\Rightarrow|a-3b+9|=|3a+b-3|$

1) $a-3b+9=3a+b-3\Rightarrow a=6-2b$

Then $C(6-2b,b), \ r=\frac{5|3-b|}{\sqrt{10}}$

and the equation is $(x+2b-6)^2+(y-b)^2=\frac{5(b-3)^2}{2}, \ b\in\mathbb{R}$

2) $a-3b+9=-3a-b+3\Rightarrow b=2a+3$

Then $C(a,2a+3), \ r=\frac{5|a|}{\sqrt{10}}$

and the equation is $(x-a)^2+(y-2a-3)^2=\frac{5a^2}{2}, \ a\in\mathbb{R}$