1. two circles. x^2+y^2-16x-8y+35=0 and x^2+y^2=5 are given.
a. write the equations in the standard form.
b.prove that the circles touch each other.
c.hence calculate the equation of their common tangent
2. find the equation of a. circle. with centre c(2;3) and which I is the tangent to the x-axis
b. find the equation of the circle which. is tangent to the x-axis at (3;0) and which passes through (1;2)
X2+y2-16x-8y+35=0
x2-16x+y2-8y=-35
x2-16x+8^2+y2-8y+4^2=-35
(x-8)^2+(y-4)^2=35
(8;4)=5/7
1. a. To write the equations in standard form, we need to expand and simplify them:
For x^2 + y^2 - 16x - 8y + 35 = 0:
Rearrange the terms: x^2 - 16x + y^2 - 8y + 35 = 0
Complete the square for x: (x^2 - 16x + 64) + y^2 - 8y + 35 = 64
Complete the square for y: (x^2 - 16x + 64) + (y^2 - 8y + 16) + 35 = 64 + 16
Simplify: (x - 8)^2 + (y - 4)^2 = 45
For x^2 + y^2 = 5:
This equation is already in standard form.
b. To prove that the circles touch each other, we need to compare the distances between their centers and the sum of their radii.
The center of the first circle is (8, 4) and the center of the second circle is the origin (0, 0).
The distance between their centers is given by √((x2 - x1)^2 + (y2 - y1)^2), which is √((0 - 8)^2 + (0 - 4)^2) = √(64 + 16) = √80 = 4√5.
The radius of the second circle is √5. The radius of the first circle can be found by taking the square root of the constant term in its standard form equation, which is √45.
Since the sum of the radii (√5 + √45) is equal to the distance between the centers (4√5), we can conclude that the circles touch each other externally.
c. The equation of the common tangent can be found by using the perpendicular bisector of the line segment connecting the centers of the circles.
The midpoint of the line segment connecting the centers is ((8 + 0)/2, (4 + 0)/2) = (4, 2).
The slope of the line connecting the centers is (4 - 2)/(8 - 0) = 1/4.
The slope of the common tangent, which is perpendicular to the line connecting the centers, is -4 (negative reciprocal of 1/4).
Using the point-slope form, we can write the equation of the common tangent as y - 2 = -4(x - 4), which simplifies to y = -4x + 18.
2. a. To find the equation of a circle with center C(2, 3) that is tangent to the x-axis, we need to find the radius.
The center of the circle is at (2, 3), and it is tangent to the x-axis. This means the distance from the center to the x-axis is equal to the radius.
The distance from the center (2, 3) to the x-axis is |3 - 0| = 3.
Therefore, the equation of the circle can be written as (x - 2)^2 + (y - 3)^2 = 3^2, which simplifies to (x - 2)^2 + (y - 3)^2 = 9.
b. To find the equation of the circle that is tangent to the x-axis at (3, 0) and passes through (1, 2), we can find the radius and the center of the circle.
The radius is the distance from the center of the circle to the x-axis, which is the y-coordinate of the center.
Since the circle is tangent to the x-axis at (3, 0), the y-coordinate of the center is 0.
The distance from the center (0, 0) to the point (1, 2) is √((1 - 0)^2 + (2 - 0)^2) = √5.
Therefore, the center of the circle is at (0, 0) and the radius is √5.
The equation of the circle can be written as x^2 + (y - 0)^2 = (√5)^2, which simplifies to x^2 + y^2 = 5.
1. a. To write the equations of the circles in standard form, we'll need to complete the square for both equations.
For the first equation, x^2 + y^2 - 16x - 8y + 35 = 0:
Rearrange the equation by grouping the x-terms and y-terms separately:
(x^2 - 16x) + (y^2 - 8y) + 35 = 0
Now, complete the square for each group:
(x^2 - 16x + 64) + (y^2 - 8y + 16) + 35 = 64 + 16
Simplifying, we get:
(x - 8)^2 + (y - 4)^2 = 45
The standard form of the first circle equation is:
(x - 8)^2 + (y - 4)^2 = 45
For the second equation, x^2 + y^2 = 5:
This equation is already in standard form.
b. To prove that the circles touch each other, we need to find the distance between their centers.
The center of the first circle is at (8, 4), and its radius is sqrt(45).
The center of the second circle is at the origin, (0, 0), and its radius is sqrt(5).
Using the distance formula, the distance between the centers is:
d = sqrt((8 - 0)^2 + (4 - 0)^2) = sqrt(64 + 16) = sqrt(80) = 4 * sqrt(5)
The sum of the radii of the two circles is:
sqrt(45) + sqrt(5)
Since the distance between the centers is equal to the sum of the radii, the circles touch each other externally.
c. To calculate the equation of the common tangent of the circles, we'll find the slope of the line connecting their centers and then use it to find the equation of the line.
The slope of the line connecting the centers is:
m = (4 - 0) / (8 - 0) = 1/2
The slope of any line perpendicular to this is -2 (negative reciprocal).
Using the point-slope form, the equation of the line is:
y - 4 = -2(x - 8)
y - 4 = -2x + 16
2x + y = 20
Therefore, the equation of the common tangent of the circles is:
2x + y = 20
2. a. To find the equation of a circle with center C(2, 3) tangent to the x-axis, we know that the radius is the distance from the center to the x-axis, which is 3.
The equation of the circle can be written as:
(x - 2)^2 + (y - 3)^2 = 3^2
Simplifying, we have:
(x - 2)^2 + (y - 3)^2 = 9
b. To find the equation of a circle that is tangent to the x-axis at (3, 0) and passes through (1, 2), we first need to find the center and radius of the circle.
Since the circle is tangent to the x-axis at (3, 0), the center lies on a line perpendicular to the x-axis passing through (3, 0). This line is a vertical line with an equation x = 3.
Since the circle passes through (1, 2), the center also lies on the perpendicular bisector of the line segment joining (1, 2) and (3, 0).
The midpoint of the line segment is:
((1 + 3) / 2, (2 + 0) / 2) = (2, 1)
Therefore, the center of the circle is (2, 1).
To find the radius, we need to find the distance between the center (2, 1) and the point (3, 0), which is:
sqrt((3 - 2)^2 + (0 - 1)^2) = sqrt(2)
The equation of the circle can be written as:
(x - 2)^2 + (y - 1)^2 = (sqrt(2))^2
(x - 2)^2 + (y - 1)^2 = 2