Question

Find the coordinates of the centre radius of each of the following circle:
(i) x² + y² + 6x – 8y – 24 = 0

(ii) 2x² + 2y² – 3x + 5y = 7

(iii) 1/2(x² + y²) + x cos θ + y sin θ - 4 = 0

(iv) x² + y² – ax – by = 0

Answer 1

(i) x² + y² + 6x – 8y – 24 = 0

Given:

The equation of the circle is x² + y² + 6x – 8y – 24 = 0 …… (1)

We know that for a circle x² + y² + 2ax + 2by + c = 0 …… (2)

Centre = (-a, -b)

So by comparing equation (1) and (2)

Centre = (-6/2, -(-8)/2)

= (-3, 4)

Radius = √(a² + b² – c)

= √(3² + 4² – (-24))

= √(9 + 16 + 24)

= √(49)

= 7

∴ The centre of the circle is (-3, 4) and the radius is 7.

(ii) 2x² + 2y² – 3x + 5y = 7

Given:

The equation of the circle is 2x² + 2y² – 3x + 5y = 7 (divide by 2 we get)

x² + y² – 3x/2 + 5y/2 = 7/2

Now, by comparing with the equation x² + y² + 2ax + 2by + c = 0 

Centre = (-a, -b)

(iii) 1/2(x² + y²) + x cos θ + y sin θ - 4 = 0

Given:

The equation of the circle is

1/2(x² + y²) + x cos θ + y sin θ - 4 = 0

(Multiply by 2 we get)

x² + y² + 2x cos θ + 2y sin θ – 8 = 0

By comparing with the equation x² + y² + 2ax + 2by + c = 0

Centre = (-a, -b)

= [(-2cos θ)/2 , (-2sin θ)/2]

= (-cos θ, -sin θ)

Radius = √(a² + b² – c)

= √[(-cos θ)² + (sin θ)² – (-8)]

= √[cos² θ + sin² θ + 8]

= √[1 + 8]

= √[9]

= 3

∴ The centre and radius of the circle is (-cos θ, -sin θ) and 3.

(iv) x² + y² – ax – by = 0

Given:

Equation of the circle is x² + y² – ax – by = 0

By comparing with the equation x² + y² + 2ax + 2by + c = 0

Centre = (-a, -b)

= (-(-a)/2, -(-b)/2)

= (a/2, b/2)

Radius = √(a² + b² – c)

= √[(a/2)² + (b/2)²]

= √[(a²/4 + b²/4)]

= √[(a² + b²)/4]

= [√(a² + b²)]/2

∴ The centre and radius of the circle is (a/2, b/2) and [√(a² + b²)]/2