Range of $k$ in $x^2+y^2-6x-10y+k=0$

Question

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 ?

Answer 1

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 $$9<k<29.$$

Answer 2

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 $9<k<29$.

Answer 3

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 $29>k>9$