If the circle x^2+y^2-2gx+6y-19c=0, g,c R passes through the point (6,1) and its centre li…

Question

If the circle $x^2+y^2-2gx+6y-19c=0$, $g,c\in\mathbb R$ 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 the $x$-axis is

Accepted Answer

Through $(6,1)$: $36+1-12g+6-19c=0\Rightarrow 12g+19c=43$.

Centre $(g,-3)$ on $x-2cy=8$: $g+6c=8$.

Solving: $c=1,\ g=2$. Circle: $x^2+y^2-4x+6y-19=0$.

$$\text{Intercept}=2\sqrt{g^2-c_{\text{const}}}=2\sqrt{4+19}=2\sqrt{23}.$$

Solutions · 1