Find the center (`h, k` ) and radius `r ` of the circle with the equation `4x^2 + 4y^2 - 12x + 16y - 5 = 0`

### 2 Answers | Add Yours

The equation of the circle is 4x^2+4y^2-12x+16y-5=0. Write this in the standard form of a circle with center (h, k) and radius r: (x - h)^2 + (y - k)^2 = r^2

4x^2+4y^2-12x+16y-5=0

=> x^2 + y^2 - 3x + 4y - 5/4 = 0

=> x^2 - 3x + 9/4 + y^2 + 4y + 4 = 5/4 + 9/4 + 4

=> (x - 3/2)^2 + (y + 2)^2 = 15/2

**The center of the circle is `(3/2, -2)` and the radius is **`sqrt 7.5`

Find the center (h, k) and radius r of the circle with the equation 4x^2 + 4y^2 - 12x + 16y - 5 = 0

4(x^2y^2-3x+4y-5/4)=0

x^2+y^2-3x+4y-5/4=0

(x^2 -2*3/2x +9/4-9/4) + (y^2 +2*2y+2^2 -4) -5/4=0

(x-3/2)^2 +(y+2)^2= 9/4 +4+5/4

(x-3/2)^2 +(y+2)^2=7.5

comparing with equation of the circle (x-h)^2 + (y-k)^=r^2,

centre(3/2,-2) and radious(r) is the square root of 7.5

**Sources:**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes