What are the coordinates of the center of a circle whose equation is (x + 7) 2 + (y – 5) 2 = 16?
What are the coordinates of the center of a circle whose equation is (x + 7) 2 + (y – 5) 2 = 16?The center is at (-7, 5)
The equation is a circle. What is the y-coordinate of the center of the circle?
y = -5
The equation for a circle is x^2 + y^2 + 10y - 7 = 0. What is the center of the circle?
You just need to know the GENERAL equation of a circle and his MEAN. If you have a point (a,b) in the plane, and you have a circle of radius R cetered in this point, then the equation of this circle is: (x-a)^2+(y-b)^2=R^2 where a, b are the coordinates x and y of the center and R. This equation comes form the application of the Phytagoream´s Theorem . Well. We need to put the given equaliy in the form of the GENERAL equation.here we can to use this ¨trick¨ (x+y)^2=x^2+2xy+y^2 we need to complete square for this purpose.As there is no term with x then: x^2+(y^2+10y+(10/2)^2)-(10/2)^2–7=0As you can see I add (10/2)^2 and subtract it. This is because for to complete square you need to add (-b/2a)^2 and subtract it. WHERE b and a are the term of: ax^2+bx or ay^2+by. As you can see b=10 and a=1Because (y+5)=y^2+10y+25.Then, resolving: x^2+y^2+10y+25–25–7=0 → x^2+(y+5)^2=32→ (x-0)^2+(y+5)^2=((32)^(1/2))^2——→ (x-0)^2+(y-(-5))^2=((32)^(1/2))^2 We can see then that a=0 and b=-5 THEN the center of circle is (0,-5) and the radius is (32)^(1/2)=4(2)^(1/2). The trick is complete square and put as the general equation of the circle.