C1→(3,2) r1=3
C2→(−5,−6) r2=3
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/866443/original_Screenshot_2020-10-08_194829.png)
Internal point of simlitude
(I) divide
C1 and
C2 in
1:1 ratio internally
so,
I≡(−1,−2)
Let the equation of tangents passing through
I be
y+2=m(x+1)
⇒mx−y+m−2=0
Perpendicular distance from
C1 to the tangent =
3
⇒∣∣∣4m−4√1+m2∣∣∣=3⇒16(m−1)2=9(m2+1)⇒7m2−32m+7=0⇒m=16±3√237
So, equation of Transverse common tangents are
y+2=16±3√237(x+1)
Radius of both circles are same.
Direct Common tangent are parallel to
C1C2, so, D.C.T. equation will be of the form
y=x+λ (∵mC1C2=1)
Perpendicular distance = radius
⇒∣∣∣1+λ√2∣∣∣=3
⇒λ+1=±3√2
⇒λ=±3√2−1
So, equation of D.C.T. are
y=x+3√2−1 and
y=x−3√2−1