Question

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

Answer 1

Answer:

c

Step-by-step explanation:

Answer 2

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)

Verify

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