BITSAT 2005 Solved Question Paper

The centre and radius of the first circle x2+y2+2x+8y−23=0 are C1(−1,−4) and r1=√40
Similarly, the centre and radius of second circle x2+y2−4x−10y+9=0 are C2(2,5) and r2=√20
Now, C1C2=√(2+1)2+(5+4)2
=√9+81=√90
r1+r2=√40+√20
r1−r2=√40−√20
Here, r1−r2<C1C2<r1+r2
∴ Two common tangents can be drawn.