Find the standard equation of the circle concentric with x^2+y^2-12x-10y = -12 and has a diameter of 10.
I have been trying to solve this all day, can you please help me :(
x^2+y^2-12x-10y = -12
x^2 - 12x + 36 + y^2 - 10y + 25 = -12+36+25
(x-6)^2 + (y-5)^2) = 49
new circle needs a radius of 5, so
(x-6)^2 + (y-5)^2 = 25
Please tell me what you did to work on this "all day" .
Of course, I'd be happy to help you!
To find the standard equation of the circle, we need to complete the square for both the x and y terms in the given equation.
First, let's rearrange the given equation:
x^2 + y^2 - 12x - 10y = -12
We will complete the square for the x terms. To do this, we need to add (12/2)^2 = 36 to both sides of the equation:
x^2 - 12x + 36 + y^2 - 10y = -12 + 36
Simplifying this, we get:
(x - 6)^2 + y^2 - 10y = 24
Similarly, we complete the square for the y terms. We need to add (10/2)^2 = 25 to both sides of the equation:
(x - 6)^2 + y^2 - 10y + 25 = 24 + 25
Simplifying this, we get:
(x - 6)^2 + (y - 5)^2 = 49
Now, we have the equation of a circle in standard form:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) represents the center of the circle and r represents the radius.
From our equation, we can identify that the center of the circle is (6, 5) and the radius squared is 49. Since we know that the diameter is 10, we can conclude that the radius is 5.
Therefore, the standard equation of the circle concentric with the given equation and with a diameter of 10 is:
(x - 6)^2 + (y - 5)^2 = 5^2
(x - 6)^2 + (y - 5)^2 = 25
I hope this explanation helps! Let me know if you have any further questions.