Respuesta :
Answer:
Part 1) r=5 units (see the explanation)
Part 2) [tex](x-4)^2+(y-7)^2=25[/tex]
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
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
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
[tex]\sqrt{(y-4)^{2}+(4-0)^{2}}=\sqrt{(4-4)^{2}+(y-2)^{2}}[/tex]
[tex]\sqrt{(y-4)^{2}+16}=\sqrt{(0)^{2}+(y-2)^{2}}[/tex]
squared both sides
[tex](y-4)^{2}+16=(y-2)^{2}[/tex]
solve for y
[tex]y^2-8y+16+16=y^2-4y+4[/tex]
[tex]y^2-8y+32=y^2-4y+4\\8y-4y=32-4\\4y=28\\y=7[/tex]
The coordinates of the center are (4,7)
step 2
Find the radius of circle C_1
[tex]r=\sqrt{(y-4)^{2}+(4-0)^{2}}[/tex]
substitute the value of y
[tex]r=\sqrt{(7-4)^{2}+(4-0)^{2}}[/tex]
[tex]r=\sqrt{(3)^{2}+(4)^{2}}[/tex]
[tex]r=\sqrt{25}[/tex]
[tex]r=5\ units[/tex]
Part 2)
Find the equation of the circle C, in standard form.
we know that
The equation of a circle in standard form is
[tex](x-h)^2+(y-k)^2=r^2[/tex]
where
(h,k) is the center
r is the radius
substitute the given values
[tex](x-4)^2+(y-7)^2=5^2[/tex]
[tex](x-4)^2+(y-7)^2=25[/tex]
Part 3) Another circle C2 has equation x² + y2 + 6x – 8y +9=0
Find the centre and radius of C2
we have
[tex]x^2+y^2+6x-8y+9=0[/tex]
Convert to standard form
[tex](x-h)^2+(y-k)^2=r^2[/tex]
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
[tex](x^2+6x)+(y^2-8y)=-9[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^2+6x+9)+(y^2-8y+16)=-9+9+16[/tex]
[tex](x^2+6x+9)+(y^2-8y+16)=16[/tex]
Rewrite as perfect squares
[tex](x+3)^2+(y-4)^2=16[/tex]
[tex](x+3)^2+(y-4)^2=4^2[/tex]
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
[tex]r=\sqrt{(0-4)^{2}+(-3+3)^{2}}[/tex]
[tex]r=\sqrt{16}[/tex]
[tex]r=4\ units[/tex] ----> is correct
therefore
The circle C_2 is tangent to the x-axis
