I need to write the standard form of the equation of the circle with the given center and radius. center (7, -4) and r = 5
kinda forgot what standard form was............
The standard form equation of a circle is a way to express the definition of a circle on the coordinate plane.
(x-h)^2+(y-k)^2=r^2
I am confused?
(x-7)^2+(y+4)^2=5^2
The standard equation with the centre (a,b) and the radius r is : \[\Large(x-a)^2+(y-b)^2=r^2\]
so you recognize the equation @violetanna ??
on @Noura11 's equation the a and b represent the center points and r represent the Radius. Do you get that so far?
yup
Okay
so would that be the standard form :(x-7)^2+(y+4)^2=5^2
or so I simplify it.....
simplify it
(x-7)^2+(y+4)^2=25
x^2-49+y^2+16=25
x^2+y^2-33=25
no, you can't simplify it like that
x^2+y^2=58
y not?
because you can only go up to (x-7)^2+(y+4)^2=25 in simplifying a Standard or General Form of the Circle's Equation
ok got it
okay hold on. I made a mistake, if you want the general form of equation you simplify more but if you want the standard you stick you (x-7)^2+(y+4)^2=25
okay?
I wanted standard Thank you so much!!!
Your welcome! Here is an example so you would understand it more. r = 5; (h, k) = (-5, 2) General Form: x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - r^2 = 0 x^2 + y^2 - 2(-5)x - 2(2)y + (-5)^2 + 2^2 - 5^2 = 0 x^2 + y^2 + 10x - 4y + 25 + 4 - 25 = 0 x^2 + y^2 + 10x - 4y + 4 = 0 ANS Standard Form: (x - h)^2 + (y - k)^2 = r^2 (x - -5)^2 + (y - 2)^2 = 5^2 (x + 5)^2 + (y - 2)^2 = 25 ANS
So don't worry about general now. Standard is easy!
wait so my final answer was (x-7)^2+(y+4)^2=25
is that wrong?
No, its right!
I was just giving you an example, did I confuse you???
awesome
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