Find center of circle touching circle $x^2+y^2-4x-6y-12=0$ internally at $(-1,-1)$

Question

Radius of larger circle is 5. Therefore difference in radii is 3, which is equal to distance between centres of the two circles. The centre of unknown circle is (h,k).

Also $$(h+1)^2+(k+1)^2=4$$

I am going to skip through a few long steps, and I finally arrive at the equation, $$3h+4k-3=0$$

Basically, I just subtracts the two second degree equations I got from the above conditions .

Now the answer is $(\frac 15, \frac 35)$ does satisfy this equation, so I am not wrong.

I just need another equation to solve them and arrive that final answer. How do I find it?

Answer 1

you just need to divide the line joining the centres i.e. from $(-1,-1)$ to $(2,3)$ in the ratio $2:3$. So the answer is $(\frac15,\frac35)$

Answer 2

Note that the given circle has its center at (2,3). You may establish the other equation recognizing that the slopes from the touch point (-1,-1) to both centers mathch, i.e.

$$\frac{k+1}{h+1}=\frac{3+1}{2+1}$$

or,

$$4h-3k+1=0$$

Answer 3

\begin{align} x^2+y^2-4x-6y-12&=0 \tag{1}\label{1} \end{align}

\eqref{1} is equivalent to

\begin{align} (x-2)^2+(y-3)^2 &= 5^2 \tag{2}\label{2} , \end{align}
that is, an equation of a circle with radius $R=5$ centered at $O=(2,3)$.

The center $O_1$ is then located at the distance $R_1=2$ from the tangent point $T$ toward $O$:

\begin{align} O_1&=T+R_1\cdot\,\frac{(O-T)}{|O-T|} \\ &=(-1,-1)+2\cdot\frac{(3,4)}{|(3,4)|} \\ &=(-1,-1)+\cdot(1.2,1.6) \\ &=(0.2,0.6) . \end{align}