Find the centre and radius of the circle
x^2 + y^2 - 6x +5y-8=0
Answers
Centre = (3, -5/2)
Radius = 93/4
Given:
Equation of circle: x²+y²-6x+5y-8=0
To find:
The center and radius of the circle
Solution:
The center and radius of the circle are (3, -5/2) and √93/2, respectively.
We can find the solution by following the given steps-
We know that the equation of a circle comprises the coordinates of the center of the circle and the value of its radius.
Let the coordinates of the center be (g, h) and the radius is r.
So, the general equation of a circle will be as follows-
(x-g)²+(y-h)²=r²
The equation given is x²+y²-6x+5y-8=0
We will convert it into the general form to determine the center and radius.
x²+y²-6x+5y-8=0
We will add 9 on both sides of the equation,
x²-6x+9+y²+5y-8=0+9
x²-6x+9+y²+5y=8+9
(x-3)²+y²+5y=17
Now, we will add 25/4 on both sides of the equation
(x-3)²+y²+5y+25/4=17+25/4
(x-3)²+(y+5/2)²=(17×4+25)/4
(x-3)²+(y+5/2)²=(17×4+25)/4
(x-3)²+(y+5/2)²=93/4
On comparing the above equation with the general equation of a circle, we can observe the following-
The coordinates of the center, (g, h)=(3, -5/2)
The value of radius²=93/4
If r²=93/4,
Radius, r=√93/2
Therefore, the center and radius of the circle are (3, -5/2) and √93/2, respectively.