Maths help, points will be awarded. Maths, circles.?

Question

a) By completing the square find the centre and radius of the circle x^2 +y^2 +4x +4 = 10y  

b) Show that (1,1) lies on the circle and find the gradient of the radius passing through this point. 

c) find the equation of the tangent to the circle at this point 

Answer 1

a) The circle equation x² + y² + 4x + 4 = 10y

The standard form of the circle equation is (x - h)² + (y - k)² = r², where, (h, k) is the center of the circle, and r is the radius.

(x² + 4x) + (y² - 10y) = - 4

To change the expression into a perfect square  add (half the x coefficient)² and add (half the y coefficient)²to each side of the expression.

Here, x coefficient = 4, so, (half of the x coefficient)² = (4/2)²= 4.

Here, y coefficient = - 10, so, (half of the y coefficient)² = (-10/2)²= 25.

Add 4 and 25 to each side.

x² + 4x + 4 + y² - 10y + 25 = - 4 + 4 + 25

(x + 2)² + (y - 5)² = 25

[x - (- 2)]² + (y - 5)² = 5²

Compare the equation with standard form of a circle equation.

The center (h, k) is (- 2, 5), and

The radius (r) is 5 units.

Answer 2

b) The circle equation in standard form [x - (- 2)]² + (y - 5)² = 5²

Substitute for(x ,y) = (1, 1) in above equation.

[1 - (- 2)]² + (1 - 5)² = 5²

9 + 16 = 25

25 = 25

The above statement is true, so (1, 1) lies on the circle.

 

To find the gradient of radius calculate the slope between center and the point.

Center (- 2, 5) = (x₁, y₁) and (x₂, y₂) = (1, 1).

Slope (m) = [(y₂ - y₁)/(x₂ -x₁)]

m = [(1 - 5)/(1 - (- 2))]

m  = - 4/3

The gradient of the radius m = - 4/3.

 

Answer 3

c) The tangent to the circle is perpendicular to the radius.

We know that perpendicular lines slopes are negative reciprocal to each other.

The tangent point (1, 1) and it's slope 3/4.

Substitute m = 3/4 and (x ,y) = (1, 1) in y = mx + b

1 = 3/4(1) + b

b = 1 - (3/4)

b = 1/4

Substitute m = 3/4 and b = 1/4 in y = mx + b.

Equation of tangent line for the circle at (1, 1) is y = (3/4)x + (1/4).