Question 283833
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:


{{{graph(400,400,-10,10,-10,10,sqrt(-(x-5)^2 + 16) - 2,-sqrt(-(x-5)^2 + 16) - 2,-2)}}}


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.