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@hartnn @waterineyes I have no clue about this one........

that is simple.

mid-point is centre

Find the distance between and divide it by 2 to find radius..

distance between the two points = diameter =2r

Do you know how to find midpoint or not??

I know nothing lol but i am going to give you a tutorial on Conic section soon promise :D

@waterineyes noooooooo

u don't know how to find mid-point ? thats news...
just take the average of co-ordinates.

x+y/2 but x1 or x2

for x-coordinate of midpoint, u take average of x-coordinates of endpoints

(-3+5)/2

Sorry net stopped working..

\[Midpoint(x,y) = (\frac{x_1 +x_2}{2}, \frac{y_1+y_2}{2})\]

@waterineyes I cannot see latex

so whats the midpoint ? centre

(x,y)=(1,-2)

thats correct centre

now find radius

Just refresh your page then..

So i use the old radius formula ?

\[r=\sqrt{g^2+f^2-c}\]

@waterineyes @hartnn Do you have a better way ?

not that, can u find distance between two points ?

What are you doing man??
Use distance formula..

Yeah

Distance between points (x1,y1) and (x2,y2) is
\(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

\[Distance = \sqrt{(x_2-x_1)^2 - (y_2-y_1)^2}\]

\[d=\sqrt{(x2-x1)^2+(y2-y1)^2}\]

lol

Written three times but used not atleast once..
ha ha ha..

lol :P

\[d=8\sqrt{2}\]

what next ?

radius = d/2

commented

This you have got is length of Diameter..
You have to find radius so you will half it..

r=4root(2)

yup... u know what to do from now....

|dw:1353175013593:dw|

(x-a)^2+(y-b)^2=r^2

Yep Go Ahead..

x^2+y^2-2x+4y-27=0

Yep..

1 + 4 = 32
implies -27 on LHS..

Thanks a lot @hartnn and @waterineyes

You are welcome dear..