AP EAMCET Engineering 2011 Solved Paper

Given that circle x2+y2+8x−4y+c=0
touch the circle x2+y2+2x+4y−11=0
then, C1C2=r1+r2 .....(i)
where C1=(−4,2)
r1=√16+4−c=√20−c
C2=(−1,−2)
and r2=√1+4+11=4
∴ From Eq. (i),
√(−4+1)2+(2+2)2
=√20−c+4
⇒ 5=√20−c+4
⇒ c=19
Also, the circle x2+y2+8x−4y+c=0
cuts the circles x2+y2−6x+8y+k=0
orthogonally, then c1→(−4,2)
C3→(3,−4)
(C1C3)2=(r1)2+(r3)2‌‌‌ where,
r1=√16+4−c
r3=√9+16−k
⇒ {(−4−3)2+(2+4)2} =(20−c)+(25−k)
⇒ 49+36=45−k−c
⇒ k+c=−40
⇒ k+19=−40
⇒ k=−59