**Rectangular Coordinates**

**Sol A** The circle has an equation of the form
or where is the distance from to the origin;
so
and the circle has the
equation .

**Sol B** The midpoint of the line segment AB is given by
, so the circle has the equation
.

**Sol 1** The midpoint of the line segment AB is given by
, so the distance from P to M is equal to
.

**Sol 2** The radius of the circle is the distance from C to the x-axis, so
. Therefore the circle has equation
.

**Sol 3** The circle has an equation of the form
, where
the radius is the distance from C to P. Instead of using the distance
formula, though, we can substitute the coordinates of P into the equation to get
or
.

**Sol 4** The radius of the circle is the distance from C to the line ,
so . Therefore the circle has equation
.

**Sol 5** The center C of the circle is the midpoint of the line segment
between P and Q, so
. Therefore the circle has
an equation of the form
, where is the distance from
C to P (or Q). Instead of finding first, though, we can substitute the
coordinates of P into this equation to get
or
.

**Sol 6** First we can find the center and radius of the circle
by completing the square:
gives
, so the circle has center and radius .
The distance from P to C is given by
. Since , the point P is inside the circle.

**Sol 7** The center of the circle is the point , and the slope of the
line through P and C is given by
. Therefore the line
through P and C has the equation
or , and the point on
the circle closest to P will be one of the points of intersection of this line
with the circle.
Substituting into the equation of the circle gives
, so
or . Then
, so and or . Since the x-coordinate of
P is 1, the point on the circle closest to P has and ; so
it is the point .

**Sol 8** First we will find an equation of the line through P which is
perpendicular to the given line. Solving for gives
, so the given line has slope and therefore a line
perpendicular to this line will have slope
.

Thus the line through P perpendicular to the given line has equation or . These two lines will intersect at a point Q which is the point on the given line closest to P, and we can find the coordinates of Q by substituting into the equation and then solving:

gives or , so and . Therefore Q is the point ,

and the distance from P to the line is
the distance from P to Q, which is given by
.

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