Question

Find the equation of a circle with
(i) centre at origin and radius 4.
(ii) centre at (3,-2)and radius 5.
(iii) centre at (-3, -2) and radius 6.
(iv) centre at (a, b) and radius $\sqrt{a^{2}+b^{2}}$.
(v) centre at (a cos α, b sin α) and radius 'a'.

Answer 1

1
centre 0,0
and radius 4
so eq= x^2+y^2=16

2 centre (3,-2)
eq= (x-3)^2+(y+2)^2=25
3 same as above
4. "
5 (x-acosa)^2+(y-bsina)^2=a^2

Answer 2

Answer:

Step-by-step explanation:

Equation of circle

( x -a)² + (y -b)² = r²

a & b are center point of circle & r = radius

(i) centre at origin and radius 4.

x² + y² = 4²

=> x² + y² = 16

(ii) centre at (3,-2)and radius 5.

(x-3)² + (y+2)² = 5²

=> x² + 9 - 6x + y² + 4 + 4y = 25

=> x² + y² -  6x + 4y - 12 = 0

(iii) centre at (-3,-2)and radius 6.

(x+3)² + (y+2)² = 5²

=> x² + 9 + 6x + y² + 4 + 4y = 36

=> x² + y² +  6x + 4y - 23 = 0

(iv) centre at (a, b) and radius√a² + b²

(x -a)² + (y-b)² = (√a² + b²)²

=> x² + a² - 2ax  + y² + b² - 2by = a² + b²

=> x² + y² - 2ax - 2ay = 0

(v) centre at (a cos α, b sin α) and radius 'a'.

(x -acosα)² + (y-bSinα)² = a²

=> x² + a²Cos²α - 2axCosα  + y² + b²Sin²α - 2bySinα = a²

=> x² + y² - a²Sin²α + b²Sin²α  - 2axCosα  - 2bySinα = 0

=> x² + y² +(b² - a²)Sin²α - 2axCosα  - 2bySinα = 0