The circle $x^2+y^2-6x-10y+k=0$ does not touch or intersect the x-axis and the point $(1,4)$ lies inside the circle, then find the range of $k$
$$ C(3,5)\;\&\;r=\sqrt{34-k} $$ Attempt 1 $$ d=\frac{|ax+by+c|}{\sqrt{a^2+b^2}}=|3|>r=\sqrt{34-k}\implies k>25 $$
Attempt 2 $$ y=0\implies x^2-6x+k=0\implies\Delta=36-4k<0 \implies k>9 $$ The solution given in my reference is $9<k<29$. The upper limit is fine, but which case is missing in attempt 1 that I am not getting the lowest limit for $k$ there ?
The distance from center $C(3,5)$ to the $x-$ axis is $d=5$ .
It must be greater than the radius. thus
$$5 >\sqrt{34-k}$$
the distance from the center to the point $(1,4)$ is
$D=\sqrt{(3-1)^2+(5-4)^2}=\sqrt{5}$
It must be smaller than the radius. thus $$\sqrt{5}<\sqrt{34-k}$$
So $$5<34-k<25$$ or $$9<k<29.$$
It does not touch x-axis means that there exists no value when y=0
Plug this in and u'll get
$$c^2-6x+k=0$$
This equation will have roots when discriminant is greater than or equal to 0
So we have $36-4k<0$ $$9<k$$
We also know that k= 34-r where r is the radius
Next we have that (1,4) is inside that means that it must satisfy $(x-3)^2+(y-5)^2 < r$
Replace r and plugin the co-ordinates
You get $(1-3)^2+(4-5)^2<34-k$
$4+1-34<-k$
So $29>k>9$
Since the point must lie inside the circle, we need $r >$ distance beteewn the point and the centre that is
$$r=\sqrt{34-k}>\sqrt{(3-1)^2+(5-4)^2}=\sqrt 5 \implies k<29$$
and therefore $9<k<29$.
