If the circle \[x^{2}+y^{2}-2 g x+6 y-19 c=0, g, c \in \mathbb{R}\] passes through the point (6,1) and its centre lies on the line x−2cy=8, then the length of intercept made by the circle on x-axis is :
Solutions · 1
Official solution
Circle : x2+y2−2gx+6y−19c=0
It passes through h(6,1)
\[ \Rightarrow 36 + 1 - 12g + 6 - 19c = 0\]
=12g+19c=43 ..... (1)
Line x−2cy=8 passes through centre
\[ \Rightarrow g + 6c = 8\] ...... (2)
From (1) \& (2)
\[g = 2,\,c = 1\]
C:x2+y2−4x+6y−19=0
x intercept \[= 2\sqrt {{g^2} - C} \]
\[ = 2\sqrt {4 + 19} \]
\[ = 2\sqrt {23} \]