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`        Find the equation of common tangents to the  circles x^2+y^2-12x-8y+36=0 and x^2+y^2-4x-2y+4=0 touching  the circles at distinct points`
2 years ago

```							Given x^2+y^2-12x-8y+36=0,x^2+y^2-4x-2y+4=0,,,C1 =(6,4) r1=4,;C2=(2,1) r2=1. C1C2=5,r1+r2=5 implies that the two circles are touch each other externally then we have two common tangents one is transverse and two direct .Now the transverse common tangent is S1-S2=0 implies 4x+3y--16=0  for the direct common tangent,let P be the external center of similitude which dibides extranally in the ratio 4:1 . there four P(2/3,0). the equation of tangent is y=m(x-2/3) which touches the circle(2) then r=perpendicular distance 1=|6m-3-2m|/√9m^2+9 implies that m=0;m=24/7 therefore the direct common tangents are y=0;24x-7y-16=0
```
2 years ago
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