Straight line x + y = 18 and common tangents of the curve S 1 = x 2 + y 2 – 28x + 160 = 0 and S 2 = 
+ 
– 1 = 0 form a triangle ABC in the first quadrant and S 3 = 0 is circumcircle of Δ ABC.
(i) Coordinates of centre of circle inscribed in the triangle ABC are –
, 10 –2 
] x 2 + y 2 – 16x – 20y + 156 = 0 
, 10 – 
] x 2 + y 2 – 10x – 8y + 156 = 0 
, 10 + 2 
] x 2 + y 2 – 8x – 10y + 156 = 0 
, 10 + 
] (ii) Equation of S 3 is – x 2 + y 2 – 20x – 16y + 156 = 0 (iii) Length of common chord of circle S 1 and S 2 is – Does not exist
Text Solution
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Ans.
(i)
Sol. Area of Δ ABC = 8 sq. unit, 2s = 4 + 4 + 4 
,
r = 
= 
= 4 – 2 
coordinate of centre
(8 + 4 – 2 
, 6 + 4 – 2 
)
= 
(ii)
Sol. centre of circumcircle is (10, 8) and radius 2 
So circle x 2 + y 2 – 20x – 16y + 156 = 0
(iii)
Sol. C 1 C 2 > r 1 + r 2
Since circles S 1 and S 3 does not intersect to each other so common chord does not exist.
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