1. ## Need help

A square is inscribed in the circle x^2+y^2-2x+4y-93=0 with the sides parallel to axes of coordinates the coordinates of vertices are??
I have solved quiet a bit but need the exact and precise solution thanks

2. Originally Posted by cnpranav
I have solved quiet a bit but need the exact and precise solution

Thank you!

3. Originally Posted by stapel

Thank you!
the co ordinates of the centre are (1,-2) I assumed it to be O
O is the center of the Diagonals..!! Let the Diagonals be AC and BD..!!
Also the length of AC and BD is sq root of 99 that much I have done can u help me??

4. Originally Posted by cnpranav
the co ordinates of the centre are (1,-2)
So you completed the square and obtained the following equation:

. . . . . $(x\, -\, 1)^2\, +\, (y\, +\, 2)^2\, =\, 98$

The center is then (h, k) = (1, -2), and the radius is $r\, =\, 7\sqrt{2}$.

Since the lower-left and upper-right corners of the square lie on a line with a slope of m = 1, and since you have a point on this line (being the shared center point), you can find the equation of the straight line passing through these corners.

Then find where this line intersects with the circle. This will give you the coordinates of two of the corners. You can use the symmetry of the square to find the other two corners.

5. Originally Posted by cnpranav
the co ordinates of the centre are (1,-2) I assumed it to be O
O is the center of the Diagonals..!! Let the Diagonals be AC and BD..!!
Also the length of AC and BD is sq root of 99 that much I have done can u help me??
hi
center u calculated is correct i.e. (1.-2)
length of diagonals is wrongly calculated
u r right that center of circle is center of diagonals....since AC and BD are diagonals and passing from center of the circle therefore they are diameter
so calculate diameter
then calculate side of the square by hyp. rule
let the side u calculated be called 'a'
assume one vertex to be (x,y)
then the remaining vertexes are: (x+a,y),(x+a,y+a),(x,y+a)
now take 2 diagonal vertexes and apply mid-point rule and get the value of (x,y).

6. Hello, cnpranav!

A square is inscribed in the circle $x^2+y^2-2x+4y-93\:=\:0$
with the sides parallel to the coordinates axes.
Find the coordinates of the vertices.
Complete the square and get: . $(x-1)^2 + (y+2)^2 \:=\:98$

The center is $(1,\text{-}2)$ and the radius is $7\sqrt{2}$}

If you make a sketch, the answer is obvious (almost).
Code:
              * * *
*           *
D o - - - - - - - o A
*|         _   * |*
|       7√2 *   |
* |         *     | *
* |       o - - - + *
* |       O       | *
|               |
*|               |*
C o - - - - - - - o B
*           *
* * *

The right triangle is isosceles: 45°-45°-90°.
The hypotenuse is $7\sqrt{2}$, and its legs are $7$.

Hence, vertex $A$ is 7 units right and 7 units up from the center: . $A(8,5)$

Got it?

7. YA I got it