Find the coordinates of the point at which the circles `x^2-y^2-4x-2y+4=0` and `x^2+y^2-12 x-8y+36=0` touch each other. Also, find equations of common tangents touching the circles the distinct points.

Answer

`y = 0 and 7y -24x + 16=0`

Explanation

two circles touch each other externally, if `C_(1)C_(2) = r_(1)+r_(2)` and internally if `C_(1)C_(2)= r_(1)~r_(2)`
Given circles are `x^(2)+y^(2) -2y+4=0`,
whose centre `C_(1)(2, 1)` and radius `r_(1)=1`
and `x^(2) +y^(2)- 12x -8y +36=0`
whose centre `C_(2)(6,4)` and radius `r_(2)=4`
The distance between the centres is
`sqrt((6-2)^(2)+(4-1)^(2))=sqrt(16+9)=5`
`rArr C_(1) C_(2) =r_(1)+r_(2)`
Therefore, the circles touch each other externally and at the point of touching the point divides the line joining the two centres internally in the ratio of their radii, `1 : 4`
Therefore, `x_(1) = (1xx6+4xx2)/(1+4)=(14)/(5)`
`y_(1)= (1xx4+4xx1)/(1+4)=(8)/(5)`
Again, to determine the equation of common tangents touching the circles in distinct points, we know that, the tangents pass through a point which divides the line joining the two centres externally in the ratio of their radi, i.e. `1:4`
Therefore, `x_(2)=(1xx6-4xx2)/(1-4)=(-2
)/(-3)=(2)/(3)`
`and y_(2)=(1xx4-4xx1)/(1-4)=0`
Now, let m be the slope of the tangent and this line passing through (2/3, 0 ) is
`y-0=m(x-2//3)`
`y-mx+(2)/(3)m=0`
This is tangent to the Ist circle, if perpendicular distance from centre = radius.
`therefore(1-2m+(2//3)m)/(sqrt(1+m^(2)))=1 [therefore C_(1)=(2,1) and r_(1)=1]`
`rArr 1-(4)/(3)m=sqrt(1+m^(2))`
`rArr1+(16)/(9)m^(2)-(8)/(3)m=1+m^(2)`
`rArr (7)/(3)m^(2)-(8)/(3)m=0`
`rArr m((7)/(9)m-(8)/3)=0`
`rArr m = 0 ,m=(27)/(7)`
Hence, the equation of two tangents are
`y=0and y=(27)/(7)(x-(2)/(3))`
`rArr y = 0 and 7y -24x + 16 = 0`