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A square is inscribed in the circle x^2+...

A square is inscribed in the circle `x^2+y^2-2x+4y-93=0` with its sides parallel to the coordinate axes. The coordinates of its vertices are `(-6,-9),(-6,5),(8,-9),(8,5)` `(-6,-9),(-6,-5),(8,-9),(8,5)` `(-6,-9),(-6,5),(8,9),(8,5)` `(-6,-9),(-6,5),(8,-9),(8,-5)`

A

`(-6,-9),(-6,5),(8,-9),(8,5)`

B

`(-6,9),(-6,-5),(8,-9),(8,5)`

C

`(-6,-9),(-6,5),(8,9),(8,5)`

D

`(-6,-9),(-6,5),(8,-9),(8,-5)`

Text Solution

Verified by Experts

The correct Answer is:
1
Equation of circle is `(x-1)^(2)+(y-2)^(2)=98`
Square is inscribed in the given circle such that its sides are parallel to coordinate axes.
So, slope of PR is 1 and that of SQ is `-1`.

Thus vertices of square are
`(1+- 7 sqrt(2)cos 45^(@), -2 +- 7 sqrt(2) sin 45^(@))-=(8,5)` and `(-6,-9)`
and `(1+7sqrt(2) cos 135^(@),-2 +-7 sqrt(2) sin 135^(@))-=(8,-9)` and `(-6,5)`
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