Respuesta :
The equation of the first circle is (x + 1)^2 + (y - 1)^2 = r^2 and the equation of the second circle is (x + 1)² + (y - 1)² = 16
The equation of the perpendicular bisector
The points are given as:
P(3, 0) and Q(0, 5)
The midpoint of PQ is
Midpoint = 0.5(3 + 0, 0 + 5)
Midpoint = (1.5, 2.5)
Calculate the slope of PQ
m = (y2 - y1)/(x2 - x1)
m = (5 - 0)/(0 - 3)
m = -5/3
A line perpendicular to PQ would have a slope (n) of
n = -1/m
This gives
n = -1/(-5/3)
n = 0.6
The equation is then calculated as:
y = n(x - x1) + y1
Where
(x1, y1) = (1.5, 2.5)
So, we have:
y = 0.6(x - 1.5) + 2.5
y = 0.6x - 0.9 + 2.5
Evaluate the sum
y = 0.6x + 1.6
Hence, the equation of the perpendicular bisector of PQ is y = 0.6x + 1.6
The center of the circle
We have:
y = x + 2
Substitute y = x + 2 in y = 0.6x + 1.6
x + 2 = 0.6x + 1.6
Evaluate the like terms
0.4x = -0.4
Divide
x = -1
Substitute x = -1 in y = x + 2
y = -1 + 2
y = 1
Hence, the center of the circle is (-1, 1)
The circle equation
We have:
Center, (a, b) = (-1, 1)
Point, (x, y) = (0, 5) and (3, 0)
A circle equation is represented as:
(x - a)^2 + (y - b)^2 = r^2
Where r represents the radius.
Substitute (a, b) = (-1, 1) in (x - a)^2 + (y - b)^2 = r^2
(x + 1)^2 + (y - 1)^2 = r^2
Substitute (x, y) = (0, 5) in (x + 1)^2 + (y - 1)^2 = r^2
(0 + 1)^2 + (5 - 1)^2 = r^2
This gives
r^2 = 17
Substitute r^2 = 17 in (x + 1)^2 + (y - 1)^2 = r^2
(x + 1)^2 + (y - 1)^2 = r^2
Hence, the circle equation is (x + 1)^2 + (y - 1)^2 = r^2
The value of a and b
The equation of the second circle is
2x² + y² + ax + by - 14 = 0
Rewrite as:
2x² + ax + y² + by = 14
For x and y, we use the following assumptions
2x² + ax = 0   and  y² + by = 0
Divide through by 2
x² + 0.5ax = 0   and  y² + by = 0
Take the coefficients of x and y
k = 0.5a          k = b
Divide by 2
k/2 = 0.25a      k/2 = 0.5b
Square both sides
(k/2)² = 0.0625a²      (k/2)² = 0.25b²
Add the above to both sides of the equations
x² + 0.5ax +0.0625a² = 0.0625a²   and  y² + by + 0.25b² = 0.25b²
Express as perfect squares
(x + 0.25a)² = 0.0625a² and (y + 0.5b)² = 0.25b²
Add both equations
(x + 0.25a)² + (y + 0.5b)² = 0.0625a² + 0.25b²
So, we have:
2x² + ax + y² + by = 14 becomes
(x + 0.25a)² + (y + 0.5b)² = 0.0625a² + 0.25b²+ 14
Comparing the above equation and (x + 1)^2 + (y - 1)^2 = r^2, we have:
0.25a = 1 and 0.5b = -1
Solve for a
a = 4 and b = -2
This means that the value of a is 4 and b is -2
Show that the second circle is in the first
We have:
a = 4 and b = -2
Substitute these values in (x + 0.25a)² + (y + 0.5b)² = 0.0625a² + 0.25b²+ 14
This gives
(x + 0.25*4)² + (y - 0.5*2)² = 0.0625*4² + 0.25*(-2)²+ 14
(x + 1)² + (y - 1)² = 16
The equation of the first circle is
(x + 1)² + (y - 1)² = 17
The radii of the first and the second circles are
R = √17
r = √16
√17 is greater than √16
Since they have the same center, and the radius of the first circle exceeds the radius of the second circle, then the second circle lies inside the first circle.
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