# find the equation of the circle passing through the extremities of the diameter of the circle

find the equation of the circle passing through the extremities of the diameter of the circle

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

$x^2 +y^2 =0$

$x^2 +y^2 -6x-8y-2=0$

I cant understand what the question asks us to do.

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The middle equation is just the origin, and not a circle at all (or if you really want, it's a degenerate circle). – Arthur Oct 14 '13 at 9:01
Cn u explain it further can't understand it... – maths lover Oct 14 '13 at 9:03
The equation $x^2 + y^2 = 0$ has only one solution, and that is $x = 0, y = 0$. And as far as drawing curves go, that means you just draw a dot at the origin. The other two equations describe full circles. Following the normal rules of equation manipulation, the first one can be rewritten as $$(x + 1)^2 + (y-2)^2 = 7$$ which means that it is a circle with center at $(-1, 2)$ (the negative of the numbers next to $x$ and $y$), and radius $\sqrt 7$ (the square root of the right-hand side). (Cont.) – Arthur Oct 14 '13 at 9:14
What this means is that if you take any point on the circle I just described, take the $x$ and $y$ coordinates of that point, and put them into the equation above, the right-hand side and left-hand side turn out to be equal. Take any other point and do the same, and they won't be equal. This is the fundamental relationship between any equation in two variables and the curve it's said to describe. That being said, I have no idea what your original question asks for. It makes little sense to me. – Arthur Oct 14 '13 at 9:15
I agree with Arthur. The original question seems like nonsense, to me. Are you sure you copied it correctly?? – bubba Oct 14 '13 at 13:01

Here $B$ denotes the center of the blue circle, $A$ denotes the center of the red circle and $H$ the center of the black circle. We know the coordinates of $B$ and $A$ hence we can find the equation of green line. After that if we solve the equation of the line with the red and the blue circles we can find coordinates of $E$ and $G$. We know the diameter of the big circle is the distance between $E$ and $G$ and the center of the big circle is mid point of $EG$. Therefore we can find the equation of the big circle.