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

Results 1 to 7 of 7

- Apr 5th 2009, 12:28 PM #1

- Joined
- Jul 2008
- Posts
- 26

- Apr 5th 2009, 12:41 PM #2

- Joined
- Mar 2007
- Posts
- 1,240

- Apr 5th 2009, 12:54 PM #3

- Joined
- Jul 2008
- Posts
- 26

- Apr 5th 2009, 01:20 PM #4

- Joined
- Mar 2007
- Posts
- 1,240

So you

**completed the square**and obtained the following equation:

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

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

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.

- Apr 5th 2009, 01:31 PM #5
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).

- Apr 6th 2009, 07:13 PM #6

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 12,028
- Thanks
- 848

Hello, cnpranav!

A square is inscribed in the circle $\displaystyle x^2+y^2-2x+4y-93\:=\:0$

with the sides parallel to the coordinates axes.

Find the coordinates of the vertices.

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

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 $\displaystyle 7\sqrt{2}$, and its legs are $\displaystyle 7$.

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

Got it?

- Apr 7th 2009, 09:46 AM #7