Courses
Courses for Kids
Free study material
Offline Centres
More
Store

How do you find the centre and radius of the following circle ${x^2} + 6x + {y^2} - 2y + 6 = 0$

Last updated date: 13th Jun 2024
Total views: 372.6k
Views today: 9.72k
Verified
372.6k+ views
Hint: First we will evaluate the centre of the circle using the standard equation of circle ${(x - a)^2} + {(y - b)^2} = {r^2}$, where $(a,b)$ is the centre of the circle and $r$ is the radius of the circle. Use the midpoint formula to evaluate the centre of the circle and hence the radius of the circle.

Complete step-by-step solution:
The general equation of circle is given by ${(x - a)^2} + {(y - b)^2} = {r^2}$
Where, $(a,b)$ is the centre of the circle and $r$ is the radius of the circle.
Now we will convert the equation ${x^2} + 6x + {y^2} - 2y + 6 = 0$ into the form ${(x - a)^2} + {(y - b)^2} = {r^2}$.
$\Rightarrow {x^2} + 6x + {y^2} - 2y + 6 = 0 \\ \Rightarrow ({x^2} + 6x + 9) + {y^2} - 2y + 6 = 9 \\ \Rightarrow {(x + 3)^2} + {y^2} - 2y + 6 = 9 \\$
Now we will subtract $6$ from both the sides.
$\Rightarrow {(x + 3)^2} + {y^2} - 2y + 6 = 9 \\ \Rightarrow {(x + 3)^2} + {y^2} - 2y = 3 \\$
Now we will complete the square of the $y$ terms.
$\Rightarrow {(x + 3)^2} + {y^2} - 2y = 3 \\ \Rightarrow {(x + 3)^2} + ({y^2} - 2y + 1) = 3 + 1 \\ \Rightarrow {(x + 3)^2} + {(y - 1)^2} = 4 \\$
First, we will evaluate the centre of the circle using the standard equation of circle ${(x - a)^2} + {(y - b)^2} = {r^2}$, where $(a,b)$ is the centre of the circle and $r$ is the radius of the circle
Now we will compare and then evaluate the coordinates of the centre ${(x + 3)^2} + {(y - 1)^2} = 4$
Hence, the centre of the circle is $( - 3,1)$.
Now we substitute all these values in the general equation of circle which is given by ${(x - a)^2} + {(y - b)^2} = {r^2}$
$\Rightarrow {(x + 3)^2} + {(y - 1)^2} = {(2)^2} \\ \Rightarrow {(x + 3)^2} + {(y - 1)^2} = 4 \\$
Hence, the radius of the circle is $4\,units$.

Note: While solving for the centre of the circle, substitute the values in the midpoint formula along with their signs. While comparing values of terms with the general equation, compare along with their respective signs. Substitute values carefully in the formulas for evaluating values of terms. While evaluating the value of the coordinates pay attention to the signs of the terms and also first compare the terms.