Analytic Geometry And Calculus Mat 102 set 1

Q1. Find the Centre of the circle..

x² +y² +3x-6y-1=0

Q2.Suppose that A(10,2) and B(3,8) are two points on the xy-plane.

find the equation of the circle on AB as its diameter.

Q3. Find the equation of the of the line through A(3,2) with gradient 5

Q4.Find the distance of P(3,2) from the line 12x+5y-11=0

Q5.find lim(x approaches 0) (tan x)/x

                                         SOLUTIONS

Q1. To find the center of the circle, rewrite the given equation into standard form. The standard form of a circle's equation is (x - h)² + (y - k)² = r², where (h, k) is the center of the circle, and r is the radius.

Given equation: x² + y² + 3x - 6y - 1 = 0

Complete the square:

x² + 3x + (3/2)² + y² - 6y + (-3)² - 1 - (3/2)² - (-3)² = 0

(x + 3/2)² + (y - 3)² = 13/4

So the center of the circle is (-3/2, 3).

Q2. To find the equation of the circle with AB as diameter, use the midpoint formula and distance formula to determine the center and radius. The midpoint formula is M(x, y) = ((x₁ + x₂)/2, (y₁ + y₂)/2), and the distance formula is d = sqrt((x₂ - x₁)² + (y₂ - y₁)²).

A(10, 2), B(3, 8)

Midpoint:

M = ((10+3)/2, (2+8)/2) = (13/2, 5)

Distance (radius):

d = sqrt((3 - 10)² + (8 - 2)²) = sqrt(49 + 36) = sqrt(85)

Equation of circle:

(x - 13/2)² + (y - 5)² = 85 or X² +y² -13x -10y +4e=0

Q3. To find the equation of the line through A(3, 2) with gradient 5, use the point-slope form, which is y - y₁ = m(x - x₁), where m is the gradient, and (x₁, y₁) is the point.

y - 2 = 5(x - 3)

y = 5x - 15 + 2

y = 5x - 13

Q4. To find the distance of P(3, 2) from the line 12x + 5y - 11 = 0, use the distance formula for a point to a line: |Ax + By + C| / sqrt(A² + B²).

A = 12, B = 5, C = -11

P(3, 2)

Distance = |12(3) + 5(2) - 11| / sqrt(12² + 5²) = |36 + 10 - 11| / sqrt(144 + 25) = |35| / sqrt(169) = 35/13

Q5. To find lim(x approaches 0) (tan x)/x, apply L'hopital's rule which states that if both functions tend to 0, you can calculate limit by finding the derivative of both functions and reevaluating the limit.

lim(x → 0) (tan x)/x

Take derivatives:

d(tan x)/dx = sec²x

d(x)/dx = 1

Reevaluate the limit:

lim(x → 0) (sec²x)

sec(0) = 1

1² = 1

So the answer to lim(x → 0) (tan x)/x is 1.