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C1 is the circle of radius 1 touching the x-axis and y-axis is the another circle of radius > 1 and touching the axes as well as C1, then the radius of C2 is

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4 years ago

```							since circle c1 touch both Axis g^2=f^2=c.from radius formula 1=(g^2+f^2-c)^1/2.we get g=f=c=1.so equation of c1 is x^2+y^2+2x+2y+1=0.equation of c2 be x^2+y^2+2gx+2fy+c=0.also c2 touch both Axis g^2=f^2=c r=[g2+f2-c]we get r=g=f.and c2 touch c1 so sum of radius=distance between centers.1+r=((g-1)^2+(f-1)^2)^1/2.1+g=[(g-1)^2+(f-1)^2]. solving we get 4g=f^2+2-2f.since g=f.we replace f with g. we get g^2-6g+1=0.solving we get g=3+2√2 or 3-2√2.g=r hence r=3+2√2 or 3-2√2 since 3-2√2 is
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3 years ago
```							since circle c1 touch  Axis g^2=f^2=c.from radius formula 1=(g^2+f^2-c)^1/2.we get g=f=c=1.so equation of c1 is x^2+y^2+2x+2y+1=0.equation of c2 be x^2+y^2+2gx+2fy+c=0.also c2 touch both Axis g^2=f^2=c r=[g2+f2-c]we get r=g=f.and c2 touch c1 so sum of radius=distance between centers.1+r=((g-1)^2+(f-1)^2)^1/2.1+g=[(g-1)^2+(f-1)^2]. solving we get 4g=f^2+2-2f.since g=f.we replace f with g. we get g^2-6g+1=0.solving we get g=3+2√2 or 3-2√2.g=r hence r=3+2√2 or 3-2√2 since 3-2√2 is
```
3 years ago
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