# Is x^2 + y^2 + 3y + 4x - 8 the equation of a circle. If yes, what is the radius.

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### 2 Answers

The general equation of a circle with center (h,k) and radius r is (x - h)^2 + (y - k)^2 = r^2.

There is a typo in the equation you have provided; it should be x^2 + y^2 + 3y + 4x - 8 = 0. To test if this is the equation of a circle, let us see if it is possible to write this in the form of the standard equation given earlier.

x^2 + y^2 + 3y + 4x - 8 = 0

=> x^2 + 4x + y^2 + 3y - 8 = 0

=> x^2 + 4x + 4 + y^2 + 3y + 9/4 - 8 - 4 - 9/4 = 0

=> (x + 2)^2 + (y + 3/2)^2 = 8 + 4 + 9/4

=> (x + 2)^2 + (y + 3/2)^2 = 14.25

**This is the equation of a circle with center (-2, -1.5) and radius `sqrt(14.25)` .**

The general equation of a circle is (x - h)^2 + (y - k)^2 = r^2 where (h,k) is the center of the circle and r is the radius of the circle.

Let's see if it is possible to express the given equation in the general form.

Note that the coefficients of both x^2 and y^2 in x^2 + y^2 + 3y + 4x - 8 = 0 are the same. Therefore it is the equation of a circle.

x^2 + y^2 + 3y + 4x - 8 = 0

x^2 + 4x + y^2 + 3y = 8

x^2 + 4x + 4 + y^2 + 3y + (3/2)^2 = 8 + 4 + (3/2)^2

(x + 2)^2 + (y + 3/2)^2 = 57/4

(x + 2)^2 + (y + 3/2)^2 = `(sqrt(57/4))^2`

The radius of the circle represented by the equation x^2 + y^2 + 3y + 4x - 8 = 0 is `sqrt(57/4)`