Show that the equation x2 + y2 – 4x + 6y – 5 = 0 represents a circ

1.	Show that the equation x2 + y2 – 4x + 6y – 5 = 0 represents a circle. Find its centre and radius.
Answer» The general equation of a conic is as follows ax² + 2hxy + by² + 2gx + 2fy + c = 0 where a, b, c, f, g, h are constants For a circle, a = b and h = 0. The equation becomes: x² + y² + 2gx + 2fy + c = 0…(i) Given, x² + y² – 4x + 6y – 5 = 0 Comparing with (i) we see that the equation represents a circle with 2g = - 4 ⇒ g = - 2, 2f = 6 ⇒ f = 3 and c = - 5. Centre ( - g, - f) = { - ( - 2), - 3} = (2, - 3). Radius = \(\sqrt{g^2+f^2-c}\) = \(\sqrt{(-2)^2+3^2-(-5)}\) = \(\sqrt{4+9+5}\) = \(\sqrt{18}\) = \(3\sqrt{2}\)

Show that the equation x2 + y2 – 4x + 6y – 5 = 0 represents a circle. Find its centre and radius.

Answer»

The general equation of a conic is as follows

ax² + 2hxy + by² + 2gx + 2fy + c = 0 where a, b, c, f, g, h are constants

For a circle, a = b and h = 0.

The equation becomes:

x² + y² + 2gx + 2fy + c = 0…(i)

Given, x² + y² – 4x + 6y – 5 = 0

Comparing with (i) we see that the equation represents a circle with 2g = - 4

⇒ g = - 2,

2f = 6

⇒ f = 3 and c = - 5.

Centre ( - g, - f) = { - ( - 2), - 3} = (2, - 3).

Radius = \(\sqrt{g^2+f^2-c}\)

= \(\sqrt{(-2)^2+3^2-(-5)}\)

= \(\sqrt{4+9+5}\) = \(\sqrt{18}\) = \(3\sqrt{2}\)

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