Write the equation of the circle in standard form. Find the center, radius, intercepts, and graph the circle.

x^2+y^2+14x-12y = 76

### 1 Answer | Add Yours

`x^2+y^2+14x -12y = −76`

`x^2+14x+49-49+y^2-12y+36-36 = -76`

`(x^2+14x+49)+(y^2-12y+36)-49-36 = -76`

`(x^2+14x+49)+(y^2-12y+36)-85 = -76`

`(x^2+14x+49)+(y^2-12y+36) = -76+85`

`(x^2+14x+49)+(y^2-12y+36) = 9`

`(x+7)^2+(y-6)^2 = 3^2`

So the standard form of the circle is;

`(x+7)^2+(y-6)^2 = 3^2`

The circle with center (h,k) and radius r can be given in standard form as `(x-h)^2+(y-k)^2 = r^2`

*Center coordinates (-7,6).*

**Radius = 3**

For x intercept y = 0

`x^2+y^2+14x-12y = −76`

`x^2+14x = -76`

`x^2+14x+76 = 0`

Discriminant `= (-14)^2-4xx1xx76 = 196-304 = -108 < 0`

*So there is no x intercept.*

For y intercept x = 0

`x^2+y^2+14x -12y = −76`

`y^2-12y = -76`

`y^2-12y +76 = 0`

Discriminant `= (-(-12))^2-4xx1xx76 = 144-304 = -240 < 0`

*So there is no y intercept.*

*Circle is attached in the image.*

**Sources:**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes