Let the circle S:36x^2+36y^2-108x+120y+C=0 be such that it neither intersects nor touches …

Question

Let the circle $S:36x^2+36y^2-108x+120y+C=0$ be such that it neither intersects nor touches the coordinate axes. If the point of intersection of the lines $x-2y=4$ and $2x-y=5$ lies inside the circle $S$, then

Accepted Answer

Dividing by $36$: $x^2+y^2-3x+\frac{10}3y+\frac C{36}=0$, centre $\left(\frac32,-\frac53\right)$, radius$^2=\frac94+\frac{25}9-\frac C{36}=\frac{181-C}{36}$.

Not touching/intersecting axes: radius $<\min\left(\frac32,\frac53\right)=\frac32$:
$$\frac{181-C}{36}<\frac94\Rightarrow181-C<81\Rightarrow C>100.$$

Lines intersect at $(2,-1)$ (solving $x-2y=4,\ 2x-y=5$). This lies inside $S$:
$$\left(2-\frac32\right)^2+\left(-1+\frac53\right)^2<\frac{181-C}{36}\ \Longrightarrow\ \frac{25}{36}<\frac{181-C}{36}\ \Longrightarrow\ C<156.$$

$100<C<156.$

Solutions · 1