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 .