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Question

If tangent to the circle x2+y2=5 at (1,2) also touches the circle x2+y28x+6y+20=0 at point (h,k), then the value of h+k is

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Solution

S1:x2+y2=5
S2:x2+y28x+6y+20=0

Equation of tangent at (1,2) to S1 is
x2y5=0

Put x=2y+5 in S2, we get
(2y+5)2+y28(2y+5)+6y+20=0
5y2+10y+5=0
(y+1)2=0
y=1
x=2+5=3
So, point of contact is (3,1).
h+k=2

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