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# Using the following equation, find the center and radius: x^{2} + 2x + y^{2} + 4y = 20

**Solution:**

Given equation x^{2} + 2x + y^{2} + 4y = 20

We know that the equation of circle with centre (h, k) and radius r is given by (x - h)^{2} +(y - k)^{2} = r^{2}

In order to bring the given equation in standard form, let us complete the square and make two perfect square trinomials on LHS.

(x^{2 }+ 2x )+ (y^{2} + 4y)= 20

(x^{2} + 2x + (-1)^{2} )+ (y^{2} + 4y + (-2)^{2}) = 20+ 1+ 4

Since a^{2} + 2ab + b^{2}= (a + b)^{2}

(x + 1)^{2} + (y + 2)^{2} = 25

By comparing, we get centre (h, k) as (-1, -2) and radius r as 5.

## Using the following equation, find the center and radius: x^{2} + 2x + y^{2} + 4y = 20

**Summary:**

Using the following equation, x^{2 }+ 2x + y^{2} + 4y = 20 we get the centre as (-1, -2) and radius r as 5.

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