0

Guest

Courses New

A square is inscribed in the circle x^2+y^2-2x+4y-93=0 with its sides parallel to the co-ordinate axes. Find the co-ordinates of its vertices.

A square is inscribed in the circle x^2+y^2-2x+4y-93=0 with its sides parallel to the co-ordinate axes. Find the co-ordinates of its vertices.

Grade:12

2 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
9 years ago
Hello Student

(x-1)^2+(y+2)^2=93+1+4=98=(7*sqrt(2))
The vertices are at a distance of 7 units (Half the side) measured in
parallel with x & y axes, from the center (1, -2) of the circle.
Hence the vertex A is (1-7, -2-7) = (-6, -9)
Vertex B is (1+7, -2-7) = (8, -9)
Vertex C is (1+7, -2+7) = (8, 5)
Vertex D is (1-7, -2+7) = (-6, 5)

Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty
asd
11 Points
5 years ago
Awesome

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More