Question

Find the center (h,k) and radius of a circle along with the intercepts if any
4x^2+24x+4y2=0

Answer 1

We can simplify this expression into the form:

\[\large{(x-a)^2 + (y-b)^2 = r^2}\]

Answer 2

I am going to simplify it. If you don't understand any step, feel free to ask

Answer 3

you need to complete the square

Answer 4

\[4x^2 + 4y^2 + 24x = 0 \tag{1}\]

\[\implies 4(x^2+y^2+6x) = 0 \tag{2}\]

\[\implies (x^2 + y^2 + 6x) =0 \tag{3}\]

\[\implies (x^2+6x + 9) + y^2 - 9 = 0 \tag{4}\]

\[\implies (x+3)^2 + (y-0)^2 = 9 \tag{5}\]

Answer 5

Now these are the 5 steps. If you need help in any step, ask it :)

Answer 6

Please note that in step 5, I used this formula:

\[\large{\boxed{(u+v)^2 = u^2 + 2uv + v^2}}\]

Answer 7

Good now compare the equation in step 5 with the equation of the circle:

\[\large{(x-a)^2 + (y-b)^2 = r^2}\]

Here, (a,b) = centre of the circle;

r = radius of circle

Answer 8

Now can you tell me the centre of the circle and its radius ?

Answer 9

that dude is long gone...:)

Answer 10

Yup but I am just leaving a response so that if he wants to find out the answer he can see the comment :)

Thanks for the medal btw