In a rectangular coordinate system, let the lines be $x+y=7$, and $x-7 y=-33$, and the circle be $x^{2}+y^{2}-28 x+6 y+165=0$.

To determine a circle whose center is 5 times as far from the first line as from the second line, which circle also passes through the intersection point of the two lines and intersects the given circle at right angles.
1. Let's determine the intersection point of the two lines.

From the two equations

$$
7-y=7 y-33
$$

which simplifies to

$$
8 y=40
$$

from which we get

$$
y=5 \quad \text { and } \quad x=7-y=2
$$

Thus, the intersection point is $M(2,5)$.

2. Let's determine the geometric locus of points that are five times farther from the first line than from the second line:

$$
\frac{x+y-7}{\sqrt{1+1}}= \pm 5 \cdot \frac{x-7 y+33}{\sqrt{1+49}}
$$

which simplifies to

$$
\frac{x+y-7}{\sqrt{2}}= \pm 5 \cdot \frac{x-7 y+33}{5 \sqrt{2}}
$$

The sought geometric locus is thus the following two lines:

$$
x+y-7=x-7 y+33
$$

which simplifies to

$$
y=5
$$

and

$$
x+y-7=-x+7 y-33
$$

which simplifies to

$$
x-3 y=-13
$$

3. Let the center of the sought circle be $O(u, v)$ and its radius $r$. According to the problem:

a) The circle passes through the point $M(2,5)$, so

$$
(u-2)^{2}+(v-5)^{2}=r^{2}
$$

b) The center of the circle lies on the line (1) or (2):

$$
v=5
$$

or

$$
u-3 v=-13
$$

c) The sought circle intersects the given circle at a right angle. If we connect the intersection points of the two circles with the centers of the circles, we get a right triangle, where the legs are the radii of the two circles and the hypotenuse is the distance between the centers of the two circles.

The equation of the given circle can be written as

$$
(x-14)^{2}+(y+3)^{2}=40
$$

and thus, by Pythagoras' theorem,

$$
(u-14)^{2}+(v+3)^{2}=r^{2}+40
$$

With this, we have three equations to determine $u, v$, and $r$. From (I) and (III),

$$
(u-2)^{2}+(v-5)^{2}=(u-14)^{2}+(v+3)^{2}-40
$$

which simplifies to

$$
3 u-2 v=17
$$

Equations (IV) and (II.1) provide the common solution set of equations (I), (II.1), and (III):

$$
u=9, \quad v=5, \quad r=7
$$

while (IV) and (II.2) provide the solution set of equations (I), (II.2), and (III):

$$
u=11, \quad v=8, \quad r=\sqrt{90}
$$

Thus, there are two solutions:

$$
(x-9)^{2}+(y-5)^{2}=49 \quad \text { and } \quad(x-11)^{2}+(y-8)^{2}=90
$$

Door Ervin (Bp., VIII., Széchenyi g. IV. o. t.)