Circle 1 C has equation ${(x + 1)^2 + (y - 1)^2}$ = 121 A circle 2 C with equation ${x^2 + y^2 -4x + 6y + p = 0}$ is drawn inside 1 C . The circles have no points of contact. What is the range of values of p?
From my understanding, circles touch internally if distance ${C_1C_2 = r_1 - r_2}$
Then I need to make sure that distance ${C_1C_2 > r_1 - r_2}$
Using the distance formula, the distance between ${C_1}$ and ${C_2}$ is 5.
${\sqrt {(2 —1)^2 + (-3 -1)^2}}$ = 5
The radius of ${C_1}$ is 11
The radius of ${C_2}$ is ${\sqrt {2^2 + 3^2 - p}}$
=> ${\sqrt {13 - p}}$
Developing this I get
${5 > {\sqrt {13 - p}} - 11}$
I am not sure how to develop this any further.
Note that $C_1$ and $C_2$ are tangents in $P$ if $r_1=r_2+d((-1,1),(2,-3))$. Then, $r_2=11-\sqrt{(2+1)^2+(-3-1)^2}=11-5=6$
Then $0\le 13-p<36$. Thus $-23<p\le 13$
Editing: The main idea
First, we prove that if $C_1$ and $C_2$ are tangent, say in $P$, then their centers and $P$ are collinear.
Let $O_1$ and $O_2$ the centers. Suppose $C_2$ inside $C_1$. Then $O_2$ is inside of $C_1$. If $O_1$, $O_2$ and $P$ aren't collinear $O_2$ is not in the segment $O_1P$. Then, $\triangle O_1O_2P$ is not degenerate. Now, cause trought $P$ passes a common tangent to $C_1$ and $C_2$, say $\ell$, taking points $A,B\in\ell$ , we have $O_1P\perp AB$ and $O_2P\perp AB$. But $\angle APB=180$, and $\angle APB=\angle APO_1+\angle O_1PO_2+\angle O_2PB=180+\angle O_1PO_2$. From this $\angle O_1PO_2=0$. But $\triangle O_1PO_2$ isn't degenerated. Thus $0<\angle O_1PO_2=0$. This contradiction shows us that $O_1$, $O_2$ and $P$ are collinear.
The case $C_2$ outside $C_1$ is analogous.