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Anettt [7]
2 years ago
8

22 A circle passes through the points

Mathematics
1 answer:
Jlenok [28]2 years ago
7 0

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

<h3>The equation of the perpendicular bisector</h3>

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

<h3>The center of the circle</h3>

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)

<h3>The circle equation</h3>

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

<h3>The value of a and b</h3>

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

<h3>Show that the second circle is in the first</h3>

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.

Read more about circle equation at:

brainly.com/question/10618691

#SPJ1

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