Question

Show that the circles `x^2+y^2-10 x+4y-20=0` and `x^2+y^2+14 x-6y+22=0` touch each other. Find the coordinates of the point of contact and the equation of the common tangent at the point of contact.

Answer 1

Two circles are
`S_(1)=x^(2)+y^(2)-10x+4y-20=0` (1)
and `S_(2)=x^(2)+y^(2)+14x-6y+22=0` (2)
For the circle (1), centre is `C_(1)(5,-2)` and radius, `r_(1)=sqrt((-5)^(2)+2^(2)-(-20))=7`.
For the circle (2), centre is `C_(2)(7,-3)` and radius, `r_(2)=sqrt(7^(2)+(-3)^(2)-22)=6`
Now, `C_(1)C_(2)=sqrt((5+7)^(2)+(-2,3)^(2))=13=r_(1)+r_(2)`
Thus, two circle are touching each other externally.
Point of contact T divides `C_(1)C_(20` internally in the ration `r_(1) : r_(2)=7:6`.
`:. T=((7(-7)+6(5))/(7+6),(7(3)+6(-2))/(7+6))-=(-19//3,9//13)`
Common tangents at point of contact is radical axis, whose equation is given by `S_(1)-S_(2)=0`.
Therefore, equation of common tangent is `-24x+10y-42=0` or `12x-5y+21=0`.