Solveeit Logo
Login

Question

Question: How do you find the centre and radius of the following circle \({x^2} + 6x + {y^2} - 2y + 6 = 0\)...

How do you find the centre and radius of the following circle x2+6x+y2−2y+6=0{x^2} + 6x + {y^2} - 2y + 6 = 0

Explanation

Solution

First we will evaluate the centre of the circle using the standard equation of circle (x−a)2+(y−b)2=r2{(x - a)^2} + {(y - b)^2} = {r^2}, where (a,b)(a,b) is the centre of the circle and rr 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=r2{(x - a)^2} + {(y - b)^2} = {r^2}
Where, (a,b)(a,b) is the centre of the circle and rr is the radius of the circle.
Now we will convert the equation x2+6x+y2−2y+6=0{x^2} + 6x + {y^2} - 2y + 6 = 0 into the form (x−a)2+(y−b)2=r2{(x - a)^2} + {(y - b)^2} = {r^2}.
⇒x2+6x+y2−2y+6=0 ⇒(x2+6x+9)+y2−2y+6=9 ⇒(x+3)2+y2−2y+6=9  \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 66 from both the sides.
⇒(x+3)2+y2−2y+6=9 ⇒(x+3)2+y2−2y=3  \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 yy terms.
⇒(x+3)2+y2−2y=3 ⇒(x+3)2+(y2−2y+1)=3+1 ⇒(x+3)2+(y−1)2=4  \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=r2{(x - a)^2} + {(y - b)^2} = {r^2}, where (a,b)(a,b) is the centre of the circle and rr 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{(x + 3)^2} + {(y - 1)^2} = 4
Hence, the centre of the circle is (−3,1)( - 3,1).
Now we substitute all these values in the general equation of circle which is given by (x−a)2+(y−b)2=r2{(x - a)^2} + {(y - b)^2} = {r^2}
⇒(x+3)2+(y−1)2=(2)2 ⇒(x+3)2+(y−1)2=4  \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 units4\,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.