Question

A square is inscribed in the circle x² + y² - 6x + 8y - 103 = 0 with its sides parallel to the coordinate axes. Then the distance of the vertex of the square which is nearest to the origin is:
(A) 6 (B) √(137)
(C) √(41) (D) 13

Answer 1

Given: A square is inscribed in the circle x² + y² - 6x + 8y - 103 = 0.

To find: Then the distance of the vertex of the square which is nearest to the origin?

Solution:

• Now, we have given the circle that is x² + y² - 6x + 8y - 103 = 0, so from this we can find out the centre of the circle.

• Equation x² + y² - 6x + 8y - 103 = 0 can be written as :

             (x-3)² + (y+4)² = 128

• So the centre is (-3,4).

• And the radius is √128 = 8√2.

• Now the diagonal of the square is equal to the twice of radius.

• So, diagonal = 2 x 8√2 = 16√2.

• From diagonal we can conclude that side of the square is 16.

• As we have given that the sides of square are parallel to coordinate axis ,  so mid point of sides would be 8 cm away  from center of circle .

• So mid point of square will be :

              (-5 , -4)  & (11 , - 4)    & (3 , -12) , (3 , 4)

• And the vertex will be:

             (-5 , -12) , (-5 , 4)    &  (11 , -12) , (11 , -4)

• Now, distance of every point from the origin is

            √5² + 12² = 13

            √11² + 4² = √137

            √11² + 12² = √265

            √5² + 4² = √41

• Now from all the above distances, √41 is smallest.

Answer:

            √41 is the distance of the nearest vertex to the origin.