Write centre and diameter of the circle x2+y2+8x+2y+13=0
Answers
x^2 + y^2 - 10x + 4y + 13 = 0
This equations is not in standard form.
Standard form of the equation of a circle is:
(x-h)^2 + (y-k)^2 = r^2 where:
(h,k) are the coordinates of the center of the circle and r is the radius of the circle.
You need to convert your equation into standard form in order to solve this problem.
Move all the terms around until all the x terms are together and all the y terms are together, and the constant term is on the right side of the equation.
You get:
x^2 - 10x + y^2 + 4y = -13
Complete the squares for the x terms and y terms.
x^2 - 10x = (x-5)^2 - 25
y^2 + 4y = (y+2)^2 - 4
Your equation becomes:
(x-5)^2 - 25 + (y+2)^2 - 4 = -13
Add 25 and add 4 to both sides of this equation to get:
(x-5)^2 + (y+2)^2 = -13 + 29 which becomes:
(x-5)^2 + (y+2)^2 = 16
The center of your circle should be (x,y) = (5,-2)
The radius of your circle should be sqrt(16) = 4
Graph your circle to see if this is true.
To graph your circle, you need to solve for y.
Your equation to work with is:
(x-5)^2 + (y+2)^2 = 16
Subtract (x-5)^2 from both sides of this equation to get:
(y+2)^2 = -(x-5)^2 + 16
Take the square root of both sides of this equation to get:
y+2 = +/- sqrt(-(x-5)^2 + 16)
Subtract 2 from both sides of this equation to get
y = +/- sqrt(-(x-5)^2 + 16) - 2
You get 2 equations out of this.
They are:
y = + sqrt(-(x-5)^2 + 16) - 2
y = - sqrt(-(x-5)^2 + 16) - 2
Graph these 2 equations and you should have your circle.
The graph of these 2 equations is shown below:
I drew a horizontal line at y = -2.
Draw an imaginary vertical line at x = 5 and you'll see that the center of the circle is at (x,y) = (5,-2).
When y = -2, the edge of the circle is at x = 1, and at x = 9. Both of these are 4 units away from x = 5, proving that the radius of the circle is 4 units long.
Centre is (-4,-1), and radius is 3
therefore diameter is 4.
the eqn can be written as
(x+4)²+(y+1)²=2²
then compare with general equation.