9th grade math !!!!!

The circle Ci, intersects the y-axis at two points, one of which is (0.4).

Anuta_ua [19.1K]

Answer:

Part 1) r=5 units (see the explanation)

Part 2) $(x-4)^2+(y-7)^2=25$

Part 3) The center of the circle is (-3,4) and the radius is 4 units

Part 4) see the explanation

Step-by-step explanation:

Part 1)

step 1

Find the center of circle C_1

we know that

The distance between the center and point (0,4) is equal to the radius

The distance between the center and point (4,2) is equal to the radius

Let

(x,y) ----> the coordinates of center of the circle

Remember that

The tangent y=2 (horizontal line) to the circle is perpendicular to the radius of the circle at point (4,2)

That means ----> The segment perpendicular to the tangent is a vertical line x=4

so

The x-coordinate of the center is x=4

The coordinates of center are (4,y)

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Remember

The distance between the center (4,y) and point (0,4) is equal to the radius

The distance between the center (4,y) and point (4,2) is equal to the radius

so

substitute

$\sqrt{(y-4)^{2}+(4-0)^{2}}=\sqrt{(4-4)^{2}+(y-2)^{2}}$

$\sqrt{(y-4)^{2}+16}=\sqrt{(0)^{2}+(y-2)^{2}}$

squared both sides

$(y-4)^{2}+16=(y-2)^{2}$

solve for y

$y^2-8y+16+16=y^2-4y+4$

$y^2-8y+32=y^2-4y+4\\8y-4y=32-4\\4y=28\\y=7$

The coordinates of the center are (4,7)

step 2

Find the radius of circle C_1

$r=\sqrt{(y-4)^{2}+(4-0)^{2}}$

substitute the value of y

$r=\sqrt{(7-4)^{2}+(4-0)^{2}}$

$r=\sqrt{(3)^{2}+(4)^{2}}$

$r=\sqrt{25}$

$r=5\ units$

Part 2)

Find the equation of the circle C, in standard form.

we know that

The equation of a circle in standard form is

$(x-h)^2+(y-k)^2=r^2$

where

(h,k) is the center

r is the radius

substitute the given values

$(x-4)^2+(y-7)^2=5^2$

$(x-4)^2+(y-7)^2=25$

Part 3) Another circle C2 has equation x² + y2 + 6x – 8y +9=0

Find the centre and radius of C2

we have

$x^2+y^2+6x-8y+9=0$

Convert to standard form

$(x-h)^2+(y-k)^2=r^2$

where

(h,k) is the center

r is the radius

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^2+6x)+(y^2-8y)=-9$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^2+6x+9)+(y^2-8y+16)=-9+9+16$

$(x^2+6x+9)+(y^2-8y+16)=16$

Rewrite as perfect squares

$(x+3)^2+(y-4)^2=16$

$(x+3)^2+(y-4)^2=4^2$

therefore

The center of the circle is (-3,4) and the radius is 4 units

Part 4) Show that the circle C2 is a tangent to the x-axis

we know that

If the x-axis is tangent to the circle, then the equation of the tangent is y=0

so

The radius of the circle must be perpendicular to the tangent

That means ----> The segment perpendicular to the tangent is a vertical line The equation of the vertical line is equal to the x-coordinate of the center

so

x=-3

The circle C_2, intersects the x-axis at point (-3,0)

<em>Verify</em>

The distance between the center (-3,4) and point (-3,0) must be equal to the radius

Calculate the radius

$r=\sqrt{(0-4)^{2}+(-3+3)^{2}}$

$r=\sqrt{16}$

$r=4\ units$ ----> is correct

therefore

The circle C_2 is tangent to the x-axis

7 0

3 years ago

Read 2 more answers

Solve.<br> 5) Solve 2y + 12 < 42.<br> Show your work.

OLEGan [10]

The answer is y < 15

4 0

3 years ago

Read 2 more answers

Easiest question in the world no joke this time

Volgvan

Question #25:

One way to solve this is by multiplying the number of packages per serving by the number of guests.

1/4 * 16 = 4 packages to serve 16 guests. This means he doesn't have enough.

A second way to solve this is by diving the number of packages by the number of packages it takes to make on serving.

3 1/2 ÷ 1/4 = 14 guests he can serve. This means he doesn't have enough.

Question #26:

First, find common denominators.

2/3 = 8/12

1/4 = 3/12

Second, subtract.

8/12 - 3/12 = 5/12

Therefore, the tank is 5/12 full at the end of the trip.

Question #27:

1/2 is cut off then 1/3 is cut off. This means 3/6 was cut off to bundle newspapers, then 1/6 was cut off to make parcel. In all, 4/6 was cut off the original string. Afterwards, 2/6 was left over. So, just multiply; 2 * 3 = 6.

Therefore, the original string was as a total of 6 meters.

Best of Luck!

3 0

3 years ago

Find equations of the spheres with center(1, −1, 6)that touch the following planes.a) xy-planeb) yz-planec) xz-plane

Misha Larkins [42]

Given :

Center of sphere , C( 1 , -1 , 6 ) .

To Find :

Find equations of the spheres with center (1, −1, 6) that touch the following planes.a) xy-plane b) yz-plane c) xz-plane .

Solution :

a)

Distance of the point from xy-plane is :

d = 6 units .

So , equation of circle with center C and radius 6 units is :

$(x-1)^2+(y-(-1))^2+(z-6)^2=6^2\\\\(x-1)^2+(y+1)^2+(z-6)^2=36$

b)

Distance of point from yz-plane is :

d = 1 unit .

So , equation of circle with center C and radius 1 units is :

$(x-1)^2+(x+1)^2+(z-6)^2=1^2\\\\(x-1)^2+(x+1)^2+(z-6)^2=1$

c)

Distance of point from xz-plane is :

d = 1 unit .

So , equation of circle with center C and radius 1 units is :

$(x-1)^2+(x+1)^2+(z-6)^2=1^2\\\\(x-1)^2+(x+1)^2+(z-6)^2=1$

Hence , this is the required solution .

4 0

3 years ago

Write the equation of the line passing through the (1,-2)

AleksAgata [21]

Theres information missing in the question. There should be one more pair of coordinates

8 0

3 years ago