Let (a, b) be the center. Then the distances from (a, b) to the given points must be equal, and, if we square
those distances, we get (a - 2)
2 + (b - 4)
2 = (a + I)
2 f (b - 2)
2
, -4a + 4 - 8b + 16 = 2a + 1 - 4b + 4, 15 =
6a + 4b. Since (a, ft) also is on the line x - 3y = 8, we have a - 3b = 8. If we multiply this equation by
—6 and add the result to 6a + 4fc = 15, we obtain 22ft = -33, b = — |. Then a =1, and the center is
(I, -1). The radius is the distance between the center and (-1, 2):
Find the points of intersection (if any) of the circles x
2 + (y -4)
2 = 10 and (x — 8)
2 + y
2 = 50.
The circles are x
2 + y
2 - 8y + 6 = 0 and x
2 - 16* + y
2 + 14 = 0. Subtract the second equation from the
first: 16* - 8>> - 8 = 0, 2x-y -1 = 0, y = 2x -1. Substitute this equation for y in the second equation:
(x - 8)
2 + (2x - I)
2 = 50, 5x
2 - 20* + 15 = 0, *
2 -4* + 3 = 0,
(x - 3)(x - 1) = 0,
x = 3 or x = 1.
Hence, the points of intersection are (3,5) and (1,1).
Let x
2 + y
2 + C^x + D^y + E 1 = 0 be the equation of a circle ^, and x
2 + y
2 + C 2 x + D 2 y + E 2 = 0 the
equation of a circle
<
# 2 that intersects ^ at two points. Show that, as k varies over all real numbers ^ — 1, the
equation (x
2 + y
2 + C^x + D^y + £,) + k(x
2 + y
2 + C 2 x + D 2 y + E 2 ) = 0 yields all circles through the intersection points of <£j and ^ 2 except ^ 2 itself.
Clearly, the indicated equation yields the equation of a circle that contains the intersection points.
Conversely, given a circle <€ ^ <&, that goes through those intersection points, take a point (x a , y a ) of "£ that
does not lie on ^ and substitute x 0 for x and y 0 for y in the indicated equation. By choice of (* 0 , y 0 ) the
coefficient of k is nonzero, so we can solve for k. If we then put this value of k in the indicated equation, we
obtain an equation of a circle that is satisfied by (x 0 , y 0 ) and by the intersection points of <£, and <£ 2 . Since three
noncollinear points determine a circle, we have an equation for
( €. {Again, it is the choice of (x 0 , y 0 ) that makes
the three points noncollinear; i.e., k¥^ -1.]
Find an equation of the circle that contains the point (3,1) and passes through the points of intersection of the two
circles x
2 + y
2 -x-y-2 = Q and x
2 + y
2 + 4x - 4y - 8 = 0.
g Problem 4.25, substitute (3,1) for (x, y) in the equation (x2 + y2 - x - y - 2) + k(x2 + y2 + 4x -I Using Problem 4.25, substitute (3,1) for (x, y) in the equation (x2 + y2 - x - y - 2) + k(x2 + y2 + 4x -
4y — 8) = 0. Then 4 + Wk = 0, k = — \. So, the desired equation can be written as
Find the equation of the circle containing the point (—2,2) and passing through the points of intersection of the
two circles x
2 + y
2 + 3x - 2y - 4 = 0 and x
2 + y
2 - 2x - y - 6 = 0.
Using Problem 4.25, substitute (-2, 2) for (x, y) in the equationx2 + y2 + 3x - 2y - 4) + k(x2 + y2 - 2x -
y — 6) = 0. Then — 6 + 4k = 0, k = \. So, the desired equation is
2(*
2 + y
2 + 3x - 2y - 4) + 3(x
2 + y
2 - 2x - y - 6) = 0
Determine the locus of a point that moves so that the sum of the squares of its distances from the lines
5* + 12_y -4 = 0 and 12* - 5y + 10 = 0 is 5. [Note that the lines are perpendicular.]
Let (x, y) be the point. The distances from the two lines are
Hence,
729 = 0, the equation of a circle.
Find the locus of a point the sum of the squares of whose distances from (2, 3) and (-1, -2) is 34.
Let (x, y) be the point. Then (x -2)
2 + (y -3)
2 + (x + I)
2 + (y + 2)
2 = 34. Simplify: x
2 + y
2 - x -
y - 8 = 0, the equation of a circle.
Find the locus of a point (x, y) the square of whose distance from (-5,2) is equal to its distance from the line
5x + 12y - 26 = 0.
Simplifying, we obtain two equations 13x2 + 13y2 + 125* - 64_y + 403 = 0 and 13x2 + I3y2 + 135* - 40y +
351 = 0, both equations of circles.
CHAPTER 4
22
4.24
4.25
4.26
4.27
4.28
4.29
4.30
Simplifying, we obtain 169*2 + 169y2 + 200* - 196y -
or
and
5*2 + 5y2 - 7y - 26 = 0
or
those distances, we get (a - 2)
2 + (b - 4)
2 = (a + I)
2 f (b - 2)
2
, -4a + 4 - 8b + 16 = 2a + 1 - 4b + 4, 15 =
6a + 4b. Since (a, ft) also is on the line x - 3y = 8, we have a - 3b = 8. If we multiply this equation by
—6 and add the result to 6a + 4fc = 15, we obtain 22ft = -33, b = — |. Then a =1, and the center is
(I, -1). The radius is the distance between the center and (-1, 2):
Find the points of intersection (if any) of the circles x
2 + (y -4)
2 = 10 and (x — 8)
2 + y
2 = 50.
The circles are x
2 + y
2 - 8y + 6 = 0 and x
2 - 16* + y
2 + 14 = 0. Subtract the second equation from the
first: 16* - 8>> - 8 = 0, 2x-y -1 = 0, y = 2x -1. Substitute this equation for y in the second equation:
(x - 8)
2 + (2x - I)
2 = 50, 5x
2 - 20* + 15 = 0, *
2 -4* + 3 = 0,
(x - 3)(x - 1) = 0,
x = 3 or x = 1.
Hence, the points of intersection are (3,5) and (1,1).
Let x
2 + y
2 + C^x + D^y + E 1 = 0 be the equation of a circle ^, and x
2 + y
2 + C 2 x + D 2 y + E 2 = 0 the
equation of a circle
<
# 2 that intersects ^ at two points. Show that, as k varies over all real numbers ^ — 1, the
equation (x
2 + y
2 + C^x + D^y + £,) + k(x
2 + y
2 + C 2 x + D 2 y + E 2 ) = 0 yields all circles through the intersection points of <£j and ^ 2 except ^ 2 itself.
Clearly, the indicated equation yields the equation of a circle that contains the intersection points.
Conversely, given a circle <€ ^ <&, that goes through those intersection points, take a point (x a , y a ) of "£ that
does not lie on ^ and substitute x 0 for x and y 0 for y in the indicated equation. By choice of (* 0 , y 0 ) the
coefficient of k is nonzero, so we can solve for k. If we then put this value of k in the indicated equation, we
obtain an equation of a circle that is satisfied by (x 0 , y 0 ) and by the intersection points of <£, and <£ 2 . Since three
noncollinear points determine a circle, we have an equation for
( €. {Again, it is the choice of (x 0 , y 0 ) that makes
the three points noncollinear; i.e., k¥^ -1.]
Find an equation of the circle that contains the point (3,1) and passes through the points of intersection of the two
circles x
2 + y
2 -x-y-2 = Q and x
2 + y
2 + 4x - 4y - 8 = 0.
g Problem 4.25, substitute (3,1) for (x, y) in the equation (x2 + y2 - x - y - 2) + k(x2 + y2 + 4x -I Using Problem 4.25, substitute (3,1) for (x, y) in the equation (x2 + y2 - x - y - 2) + k(x2 + y2 + 4x -
4y — 8) = 0. Then 4 + Wk = 0, k = — \. So, the desired equation can be written as
Find the equation of the circle containing the point (—2,2) and passing through the points of intersection of the
two circles x
2 + y
2 + 3x - 2y - 4 = 0 and x
2 + y
2 - 2x - y - 6 = 0.
Using Problem 4.25, substitute (-2, 2) for (x, y) in the equationx2 + y2 + 3x - 2y - 4) + k(x2 + y2 - 2x -
y — 6) = 0. Then — 6 + 4k = 0, k = \. So, the desired equation is
2(*
2 + y
2 + 3x - 2y - 4) + 3(x
2 + y
2 - 2x - y - 6) = 0
Determine the locus of a point that moves so that the sum of the squares of its distances from the lines
5* + 12_y -4 = 0 and 12* - 5y + 10 = 0 is 5. [Note that the lines are perpendicular.]
Let (x, y) be the point. The distances from the two lines are
Hence,
729 = 0, the equation of a circle.
Find the locus of a point the sum of the squares of whose distances from (2, 3) and (-1, -2) is 34.
Let (x, y) be the point. Then (x -2)
2 + (y -3)
2 + (x + I)
2 + (y + 2)
2 = 34. Simplify: x
2 + y
2 - x -
y - 8 = 0, the equation of a circle.
Find the locus of a point (x, y) the square of whose distance from (-5,2) is equal to its distance from the line
5x + 12y - 26 = 0.
Simplifying, we obtain two equations 13x2 + 13y2 + 125* - 64_y + 403 = 0 and 13x2 + I3y2 + 135* - 40y +
351 = 0, both equations of circles.
CHAPTER 4
22
4.24
4.25
4.26
4.27
4.28
4.29
4.30
Simplifying, we obtain 169*2 + 169y2 + 200* - 196y -
or
and
5*2 + 5y2 - 7y - 26 = 0
or