Question

Radius of the circle x^2+y^2-3x-2y-3/4=0 is

Answer 1

Formula:

If r be the radius of the circle and (a, b) be the centre of the circle, then the equation of the circle is

(x - a)² + (y - b)² = r²

Solution:

The equation of the given circle is

x² + y² - 3x - 2y - 3/4 = 0

or, x² - 3x + y² - 2y = 3/4

or, x² - (2 * x * 3/2) + (3/2)² + y² - (2 * y * 1) + 1² = 3/4 + (3/2)² + 1²

or, (x - 3/2)² + (y - 1)² = 3/4 + 9/4 + 1

or, (x - 3/2)² + (y - 1)² = (3 + 9 + 4)/4

or, (x - 3/2)² + (y - 1)² = 16/4

or, (x - 3/2)² + (y - 1)² = 4

or, (x - 3/2)² + (y - 1)² = 2²

Comparing it with the equation of the circle in centre, radius form from the formula part, we get

(a, b) ≡ (3/2, 1) and r = 2 units

∴ the radius of the circle is 2 units.

Answer 2

Answer:

Radius of the circle = 2 units.

Step-by-step explanation:

As per the question,

Given equation of circle is  x² + y² - 3x - 2y - 3/4 = 0

Formula used:

If r be the radius of the circle and (a, b) be the centre of the circle, then the equation of the circle is

(x - a)² + (y - b)² = r²

Where

(a , b) is the coordinate of centre.

r = radius of the circle.

Now,

$x^{2}+y^{2}-3x-2y-\frac{3}{4} =0$

Can be written as

$(x-\frac{3}{2})^{2}+(y-1)^{2}=2^{2}$

Comparing this equation with the standard equation of the circle, we can conclude that

Centre = (a, b) = (3/2, 1)

radius = r = 2 units

Therefore, The radius of the circle is 2 units.