Finding equation of circle under the given geometric conditions 
Finding the equation of the circle which touches the pair of lines
  $7x^2 - 18xy +7 y^2 = 0$ and the circle $x^2 + y^2 - 8x -8y = 0$ and
  contained in the given circle??

My attempt
The centre of required circle would lie on angle bisector of the pair of lines ie $x=y$. 
Assuming circle to be $(x-h)^2+(y-h)^2=r^2$
Now $2(h-8)^2=r^2$ ( distance between the extreme of larger circle and center of contained circle,)
I am unable to frame a second equation . One way would be to calculate the angle between pair of straight lines and use it to find a relation between $r$ and $h$.
However I was looking for a better solution or suggestion ?
 A: You are almost there. All you need is to use the fact that the circle center is equidistant from both lines. In particular, using the distance formula, you can write
$$r^2 = \frac{((9+4\sqrt{2})h-7h)^2}{(9+4\sqrt{2})^2+7^2} = 2(h-8)^2 \implies h=6,12.$$
Since the center $(h,h)$ lies in the bigger circle, $h\neq 12$. Consequently, $h=6$ and $r^2=8.$
So $$(x-6)^2+(y-6)^2=8$$ is the equation of the sought circle.
Note that the tangent lines are $7y = (9\pm 4\sqrt{2})x.$
A: The angle bisector of the two lines $y=\dfrac{1}{7} \left(9 \pm4 \sqrt{2} \right)x$ is the line $y=x$
The wanted circle has then centre $H(k,k)$ and its radius is the distance from the given lines $\left(9+4 \sqrt{2}\right) x-7 y=0$
$$r_k=\frac{\left|\left(9+4 \sqrt{2}\right) k-7 k\right|}{\sqrt{\left(9+4 \sqrt{2}\right)^2+49}}=\frac{\left|2 \left(1+2 \sqrt{2}\right) k\right|}{12+3 \sqrt{2}}$$
The wanted circle must also be tangent to the given circle $x^2-8 x+y^2-8 y=0$ having centre $C(4,4)$ and radius $R=4\sqrt{2}$ 
A circle is internally tangent to another when the distance of the centres is equal to the difference of the radii (in absolute value)
Therefore we must have $CH=R-r_k$
that is $$\sqrt{(4-k)^2+(4-k)^2}=4\sqrt{2}-\frac{\left|2 \left(1+2 \sqrt{2}\right) k\right|}{12+3 \sqrt{2}}$$
which simplified, noticing that $k>4$ becomes
$$\sqrt{2}(k-4)=\frac{2 \left(1+2 \sqrt{2}\right) (12-k)}{3 \left(4+\sqrt{2}\right)}$$
After a look to the conditions we can say that $k>4$ so the previous equation becomes
$$\left(12+3 \sqrt{2}\right) \sqrt{2} (k-4)=2 \left(1+2 \sqrt{2}\right) k$$
Solution is $k=6$ and the wanted circle has equation
$(x-6)^2+(y-6)^2=8\to \color{red}{x^2+y^2-12x-12y+64=0}$

A: Hint:

The sought circle is the incircle of $\triangle ABC$.
