# What is the center of the circle x^2 + y^2 + 3x + 4y = 45

*print*Print*list*Cite

### 2 Answers

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

Express the equation of the circle given in the form described.

x^2 + y^2 + 3x + 4y = 45

=> x^2 + 3x + y^2 + 4y = 45

=> x^2 + 3x + (3/2)^2 + y^2 + 4y + 4 = 45 + 9/4 + 4

=> (x + 3/2)^2 + (y + 2)^2 = 205/4

**The center of the circle is (-3/2, -2)**

`A)` You can re-write the eqaution as:

`[x-2]^2+[x-(-3)]^2=5^2`

So the circle has the centre in the point C(2;-3)

`B)` to se if the point A(6-6) lies on the cirlce is enough to see if the distance d from the center C(2,-3) is R.

indeed :

`d^2=(6-2)^2+(-6+3)^2=4^2+3^2=5^2=R`

The the point lies on the circle.