a circle has equation (x-2)^2 + (y+3)^2=25

Question

a. find the centre and radius of the cirle
b.verify that the point A(6,-6) is on the circle
c. if [AB] is a diameter of the circle, find point B.
d.Find the equation of the tangent to the circle A

Answer 1

(x-2)^2 + (y+3)^2=25

The general formula of circle which have (h,k),radius(r) is

(x-h)^2 +(y-k)^2 = r^2

(x - 2)^2 + (y + 3)^2 = 5^2

(x - 2)^2  + (y -(-3))^2 =5^2

(x - h)^2 +(y - k)^2 = r^2

center (h, k) = (2,-3)

radius(r) = 5

Answer 2

b.The circle equation is (x-2)^2 + (y+3)^2=25

The point (6,-6) substituted in the circle

(x -2)^2 + (y +3)^2 = 25

(6 -2)^2 + ((-6)+3)^2  = 25

(4)^2 + (-3)^2  = 25

16 + 9 = 25

25 = 25

This is true.

The point (6,-6) lie on the circle

Answer 3

c. [AB] is a diameter of the circle.

The centre of circle is mid point of the diameter.

centre (h,k) = (2,-3)

given point  A =(6,-6)

             Let  B = (x ,y)

A(6,-6) B(x ,y) centre(h,k)=(2,-3)

(x₁,x₂)(y₁,y₂)center of circle lie on the mid point of the diameter is

(x₁ + x₂)/2 , (y₁ + y₂)/2= (h,k)

(6 + x)/2 , ((-6) + y)/2= (2 ,-3 )

(6 + x)/2 = 2 , ((-6) + y)/ = -3

6 + x  = 4 , -6 + y = -6

x = 4-6 , y = -6+6

x = -2 ,  y  = 0

point B =(-2,0)

Answer 4

d)  Circle equation is (x-2)^2 + (y+3)^2=25

Consider derivative on each side

d/dx((x-  2)^2 + (y + 3)^2 )= d/dx(25)

2(x - 2)^2-1 + 2(y + 3)^2-1dy/dx  = 0

2(x -2) + 2(y +3)dy/dx = 0

2(y + 3)dy/dx = -2(x - 2)

dy/dx = -2(x - 2)/2(y + 3)

dy/dx =- (x-2)/(y+3)

A = (6, -6)

dy/dx = -(6-2)/(-6+3)

dy/dx = 4/3

Use point slope form to find equation of the tangent line to the circle

The point slope from is y - y₁ = m(x - x₁)

y - 6 = 4/3 (x - (-6))

3(y - 6) = 4(x + 6)

3y - 18 =4x + 24

4x - 3y + 24 + 18 = 0

4x - 3y + 42 =0

Comments

Point (x₁, y₁) = (6, - 6) and slope m = 4/3.

The point-slope from of line equation is y - y₁ = m(x - x₁).

y - (- 6) = 4/3 (x - 6)

3(y + 6) = 4(x - 6)

3y + 18 = 4x - 24

4x - 3y - 24 - 18 = 0

4x - 3y - 42 = 0.

The tangent line equation is 4x - 3y - 42 = 0.