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Using the following equation, find the center and radius: x2 + 4x + y2 - 6y = -4
Solution:
Given equation x2 + 4x + y2 - 6y = -4
We know that the equation of circle with centre (h, k) and radius r is given by (x - h)2 +(y - k)2 = r2
In order to bring the given equation in standard form, let us convert the LHS into two perfect square trinomials.
We use completing the square method for this.
x2 + 4x + y2 - 6y = -4
(x2 + 4x + 22 ) + (y2 - 6y + 32) = -4 + 4 + 9
(x2 + 4x + 4 ) + (y2 - 6y + 9) = 9
[since a2 + 2ab + b2 = (a + b)2 and a2 - 2ab + b2 = (a - b)2]
(x + 2)2 + (y - 3)2 = 9
(x + 2)2 + (y - 3)2 = (3)2
By comparing, we get centre (h, k) as (-2, 3) and radius r as 3.
Using the following equation, find the center and radius: x2 + 4x + y2 - 6y = -4
Summary:
Using the following equation, x2 + 4x + y2 - 6y = -4 the centre is (-2, 3) and the radius is 3 units.
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