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In a rectangle $ABCD$, the coordinates of $A$ and $B$ are $(1,2)$ and $(3,6)$ respectively and some diameter of the circumscribing circle of $ABCD$ has equation $2x-y+4=0$. Then the area of the rectangle is:

My work: I found the equations of $AD$ and $BC$ of the rectangle. Taking the points $C$ and $D$ as $(x_1,y_1)$ and $(x_2,y_2)$ I wrote the equation for $AD=BC$. I think that the equation given for the diameter of the circle should pass through the point of intersection of the diagonals of the rectangle and wrote equation for the point of intersection. This gives me two equations and four unknowns. I know there is some problem with my method making the answer to this problem hard. So, please help me do the problem by the right method.

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Since you know it is a KVPY exam paper, can't you just do a google search for part of the problem? Turns up a lot of hits – Calvin Lin Oct 10 '13 at 21:22
Sir, It's a multiple choice question and I will get only the question and the answer key for it in the net and I know it because I had searched before. – Rajath Krishna R Oct 11 '13 at 0:44

HINT: The slope of the line $AB$ (i.e $2$) is equal to that of the diameter, that is they are parallel.

$\therefore$ The diameter must go through the mid-ponts of $BC$ and $AD$.

$\therefore$ Perpendicular distance between the diameter and $AB$ is $\frac{1}{2}BC$.

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Here is one ugly way.

Since the diameter lies on the given line, we know the centre of the circle $P=(x,y)$ lies on this line. The distance from $P$ to $A=(1,2)$ and $B=(3,6)$ must be the same so we have the equation: $(x-1)^2+(y-2)^2 = (x-3)^2+(y-6)^2$. Simplifying gives the equation $2y+x = 10$. Since $P$ lies on the given line, we must also have $2x-y+4 = 0$ as well, and solving gives the rather ugly $P=\frac{1}{5}(2, 24)$.

Now we can compute $C$ by reflecting through $P$, to get $C = P+(P-A) = \frac{1}{5}(-1, 38)$, then the area is given by $|A-B||B-C| = \sqrt{(2^2+4^2)\frac{1}{25}(16^2+8^2)} = 16$.

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Why the downvote? – copper.hat Feb 22 '14 at 16:34

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