find the centre and radius of

Question

x²+y²+4x-6y-5=0.

Answer 1

Given circleequation is x^2+y^2+4x-6y-5 = 0

Compare it to the circle equation is x^2+y^2+2gx+2fy+c = 0.

2g = 4 then g = 2

2f = -6 then f = -3

Center (-g,-f) = (-2,3)

r = √(g^2+f^2-c)

r = √(4+9+5)

r = √18

Radius of the given circle is √18.

Answer 2

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.

The equation is x² + y² + 4x - 6y - 5 = 0.

Write the equation in standard form of a circle.

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 the x coefficient)² = (4/2)²= 4.

Here, y coefficient = - 6, so, (half the y coefficient)² = (- 6/2)²= 9.

Add 4 and 9 to each side.

x² + 4x + 4 + y² - 6y + 9 - 5 = 0 + 4 + 9

(x + 2)² + (y - 3)² = 5 + 13 = 18

(x - (- 2))² + (y - 3))² = (√18)² .

Compare the equation with standard form of a circle equation.

The center (h, k) is (- 2, 3), and

The radius (r) is √18 units.