Question

How do you find the centre and the radius of the given circle: \[{x^2} + {y^2} + 6y = - 50 - 14x\] ?

Answer 1

Hint: To solve this question, use the standard equation of a circle. Simplify the given expression into the standard form using various identities, and then by comparing the obtained equation and the standard form, find the coordinates of the centre and the length of radius.

Complete step-by-step answer:
We have the following equation of circle: \[{x^2} + {y^2} + 6y = - 50 - 14x\] .
In order to find the centre and the radius of this circle, we first need to convert the equation into the standard form of the circle’s equation.
The standard form is \[{(x - h)^2} + {(y - k)^2} = {r^2}\] , where \[(h,k)\] is the centre of the circle, and \[r\] is the radius of the circle.
We will now simplify the given equation into the standard form. We have,
  \[{x^2} + {y^2} + 6y = - 50 - 14x\]
Notice that \[ - 14x\] on the right side of the equation. That can be used in the identity: \[{(a + b)^2} = {a^2} + 2ab + {b^2}\] . Thus, we will apply this identity for both the variables.
Adding \[14x\] and \[49\] to both sides of the equation, we get
  \[{x^2} + {y^2} + 14x + 6y + 49 = -1\]
On further simplification, we have
  \[{(x + 7)^2} + {y^2} + 6y = -1\]
Now, we need to perform the same procedure for the other variable.
Adding \[9\] on both sides of the equation, we get
 \[{(x + 7)^2} + {y^2} + 6y + 9 = 8\]
After simplifying, we will obtain the following expression:
  \[{(x + 7)^2} + {(y + 3)^2} = 8\]
Thus, by comparing the final equation with the original formula, we obtain the following values.
Centre: \[( - 7, - 3)\]
Radius: \[\sqrt {8} \]
So, the correct answer is “Centre: \[( - 7, - 3)\]
Radius: \[\sqrt {8} \] ”.

Note: Circle equation formula is the equation of a circle which represents the centre-radius form of the circle. A circle is referred to a round shape boundary where all the points on the boundary are equidistant from the centre. An equation is generally required to represent the circle mathematically.